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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•12–1. A car starts from rest and with constant
acceleration achieves a velocity of 15 m>s when it travels a
distance of 200 m. Determine the acceleration of the car
and the time required.
Kinematics:
v0 = 0, v = 15 m>s, s0 = 0, and s = 200 m.
+ B
A:
v2 = v0 2 + 2ac(s - s0)
152 = 02 + 2ac(200 - 0)
ac = 0.5625 m>s2
+ B
A:
Ans.
v = v0 + act
15 = 0 + 0.5625t
t = 26.7 s
Ans.
12–2. A train starts from rest at a station and travels with
a constant acceleration of 1 m>s2. Determine the velocity of
the train when t = 30 s and the distance traveled during
this time.
Kinematics:
ac = 1 m>s2, v0 = 0, s0 = 0, and t = 30 s.
+ B
A:
v = v0 + act
= 0 + 1(30) = 30 m>s
+ B
A:
s = s0 + v0t +
= 0 + 0 +
Ans.
1 2
at
2 c
1
(1) A 302 B
2
= 450 m
Ans.
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12–3. An elevator descends from rest with an acceleration
of 5 ft>s2 until it achieves a velocity of 15 ft>s. Determine the
time required and the distance traveled.
Kinematics:
ac = 5 ft>s2, v0 = 0, v = 15 ft>s, and s0 = 0.
A+TB
v = v0 + act
15 = 0 + 5t
t = 3s
A+TB
Ans.
v2 = v0 2 + 2ac(s - s0)
152 = 02 + 2(5)(s - 0)
s = 22.5 ft
Ans.
*12–4. A car is traveling at 15 m>s, when the traffic light
50 m ahead turns yellow. Determine the required constant
deceleration of the car and the time needed to stop the car
at the light.
Kinematics:
v0 = 0, s0 = 0, s = 50 m and v0 = 15 m>s.
+ B
A:
v2 = v0 2 + 2ac(s - s0)
0 = 152 + 2ac(50 - 0)
ac = -2.25 m>s2 = 2.25 m>s2 ;
+ B
A:
Ans.
v = v0 + act
0 = 15 + (-2.25)t
t = 6.67 s
Ans.
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•12–5. A particle is moving along a straight line with the
acceleration a = (12t – 3t1/2) ft>s2, where t is in seconds.
Determine the velocity and the position of the particle as a
function of time. When t = 0, v = 0 and s = 15 ft.
Velocity:
+ B
A:
dv = a dt
L0
v
dv =
L0
t
A 12t - 3t1>2 B dt
v冷0 = A 6t2 - 2t3>2 B 2
v
t
0
v = A 6t2 - 2t3>2 B ft>s
Ans.
Position: Using this result and the initial condition s = 15 ft at t = 0 s,
+ B
A:
ds = v dt
s
L15 ft
ds =
L0
t
A 6t2 - 2t3>2 B dt
s冷15 ft = a2t3 s
s = a2t3 -
4 5>2 t
t b2
5
0
4 5>2
t + 15 b ft
5
Ans.
12–6. A ball is released from the bottom of an elevator
which is traveling upward with a velocity of 6 ft>s. If the ball
strikes the bottom of the elevator shaft in 3 s, determine the
height of the elevator from the bottom of the shaft at the
instant the ball is released. Also, find the velocity of the ball
when it strikes the bottom of the shaft.
Kinematics: When the ball is released, its velocity will be the same as the elevator at
the instant of release. Thus, v0 = 6 ft>s. Also, t = 3 s, s0 = 0, s = -h, and
ac = -32.2 ft>s2.
A+cB
s = s0 + v0t +
1
a t2
2 c
-h = 0 + 6(3) +
1
(-32.2) A 32 B
2
h = 127 ft
A+cB
Ans.
v = v0 + act
v = 6 + (-32.2)(3)
= -90.6 ft>s = 90.6 ft>s
T
Ans.
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12–7. A car has an initial speed of 25 m>s and a constant
deceleration of 3 m>s2. Determine the velocity of the car
when t = 4 s. What is the displacement of the car during the
4-s time interval? How much time is needed to stop the car?
v = v0 + act
v = 25 + (-3)(4) = 13 m>s
¢s = s - s0 = v0 t +
Ans.
1
a t2
2 c
¢s = s - 0 = 25(4) +
1
(-3)(4)2 = 76 m
2
Ans.
v = v0 + ac t
0 = 25 + (-3)(t)
t = 8.33 s
Ans.
*12–8. If a particle has an initial velocity of v0 = 12 ft>s to
the right, at s0 = 0, determine its position when t = 10 s, if
a = 2 ft>s2 to the left.
+ B
A:
s = s0 + v0 t +
1
a t2
2 c
= 0 + 12(10) +
1
(-2)(10)2
2
= 20 ft
Ans.
•12–9. The acceleration of a particle traveling along a
straight line is a = k>v, where k is a constant. If s = 0,
v = v0 when t = 0, determine the velocity of the particle as
a function of time t.
Velocity:
+ B
A:
dv
a
dt =
L0
t
L0
t
t
v
dt =
v
dt =
t2 =
0
t =
dv
k>v
Lv0
1
vdv
k
Lv0
1 2 v
v 2
2k v0
1
A v2 - v0 2 B
2k
v = 22kt + v0 2
Ans.
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12–10. Car A starts from rest at t = 0 and travels along a
straight road with a constant acceleration of 6 ft>s2 until it
reaches a speed of 80 ft>s. Afterwards it maintains this
speed. Also, when t = 0, car B located 6000 ft down the
road is traveling towards A at a constant speed of 60 ft>s.
Determine the distance traveled by car A when they pass
each other.
60 ft/s
A
6000 ft
Distance Traveled: Time for car A to achives y = 80 ft>s can be obtained by
applying Eq. 12–4.
+ B
A:
y = y0 + ac t
80 = 0 + 6t
t = 13.33 s
The distance car A travels for this part of motion can be determined by applying
Eq. 12–6.
+ B
A:
y2 = y20 + 2ac (s - s0)
802 = 0 + 2(6)(s1 - 0)
s1 = 533.33 ft
For the second part of motion, car A travels with a constant velocity of y = 80 ft>s
and the distance traveled in t¿ = (t1 - 13.33) s (t1 is the total time) is
+ B
A:
s2 = yt¿ = 80(t1 - 13.33)
Car B travels in the opposite direction with a constant velocity of y = 60 ft>s and
the distance traveled in t1 is
+ B
A:
B
s3 = yt1 = 60t1
It is required that
s1 + s2 + s3 = 6000
533.33 + 80(t1 - 13.33) + 60t1 = 6000
t1 = 46.67 s
The distance traveled by car A is
sA = s1 + s2 = 533.33 + 80(46.67 - 13.33) = 3200 ft
Ans.
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12–11. A particle travels along a straight line with a
velocity v = (12 - 3t2) m>s, where t is in seconds. When
t = 1 s, the particle is located 10 m to the left of the origin.
Determine the acceleration when t = 4 s, the displacement
from t = 0 to t = 10 s, and the distance the particle travels
during this time period.
v = 12 - 3t2
(1)
dv
= -6t 2
= -24 m>s2
dt
t=4
a =
s
L-10
ds =
L1
t
v dt =
L1
t
Ans.
A 12 - 3t2 B dt
s + 10 = 12t - t3 - 11
s = 12t - t3 - 21
s| t = 0 = -21
s|t = 10 = -901
¢s = -901 - ( -21) = -880 m
Ans.
From Eq. (1):
v = 0 when t = 2s
s|t = 2 = 12(2) - (2)3 - 21 = -5
sT = (21 - 5) + (901 - 5) = 912 m
Ans.
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*12–12. A sphere is fired downwards into a medium with
an initial speed of 27 m>s. If it experiences a deceleration of
a = (-6t) m>s2, where t is in seconds, determine the
distance traveled before it stops.
Velocity: y0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have
A+TB
dy = adt
y
L27
dy =
L0
t
-6tdt
y = A 27 - 3t2 B m>s
[1]
At y = 0, from Eq.[1]
0 = 27 - 3t2
t = 3.00 s
Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result y = 27 - 3t2 and applying
Eq. 12–1, we have
A+TB
ds = ydt
L0
s
ds =
L0
t
A 27 - 3t2 B dt
s = A 27t - t3 B m
[2]
At t = 3.00 s, from Eq. [2]
s = 27(3.00) - 3.003 = 54.0 m
Ans.
•12–13. A particle travels along a straight line such that in
2 s it moves from an initial position sA = +0.5 m to a
position sB = -1.5 m. Then in another 4 s it moves from sB
to sC = +2.5 m. Determine the particle’s average velocity
and average speed during the 6-s time interval.
¢s = (sC - sA) = 2 m
sT = (0.5 + 1.5 + 1.5 + 2.5) = 6 m
t = (2 + 4) = 6 s
vavg =
¢s
2
= = 0.333 m>s
t
6
(vsp)avg =
Ans.
sT
6
= = 1 m>s
t
6
Ans.
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12–14. A particle travels along a straight-line path such
that in 4 s it moves from an initial position sA = -8 m to a
position sB = +3 m. Then in another 5 s it moves from sB to
sC = -6 m. Determine the particle’s average velocity and
average speed during the 9-s time interval.
Average Velocity: The displacement from A to C is ¢s = sC - SA = -6 - (-8)
= 2 m.
yavg =
¢s
2
=
= 0.222 m>s
¢t
4 + 5
Ans.
Average Speed: The distances traveled from A to B and B to C are
sA : B = 8 + 3 = 11.0 m and sB : C = 3 + 6 = 9.00 m, respectively. Then, the
total distance traveled is sTot = sA : B + sB : C = 11.0 - 9.00 = 20.0 m.
A ysp B avg =
sTot
20.0
=
= 2.22 m>s
¢t
4 + 5
Ans.
12–15. Tests reveal that a normal driver takes about 0.75 s
before he or she can react to a situation to avoid a collision.
It takes about 3 s for a driver having 0.1% alcohol in his
system to do the same. If such drivers are traveling on a
straight road at 30 mph (44 ft>s) and their cars can
decelerate at 2 ft>s2, determine the shortest stopping
distance d for each from the moment they see the
pedestrians. Moral: If you must drink, please don’t drive!
v1 ⫽ 44 ft/s
d
Stopping Distance: For normal driver, the car moves a distance of
d¿ = yt = 44(0.75) = 33.0 ft before he or she reacts and decelerates the car. The
stopping distance can be obtained using Eq. 12–6 with s0 = d¿ = 33.0 ft and y = 0.
+ B
A:
y2 = y20 + 2ac (s - s0)
02 = 442 + 2(-2)(d - 33.0)
d = 517 ft
Ans.
For a drunk driver, the car moves a distance of d¿ = yt = 44(3) = 132 ft before he
or she reacts and decelerates the car. The stopping distance can be obtained using
Eq. 12–6 with s0 = d¿ = 132 ft and y = 0.
+ B
A:
y2 = y20 + 2ac (s - s0)
02 = 442 + 2(-2)(d - 132)
d = 616 ft
Ans.
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*12–16. As a train accelerates uniformly it passes
successive kilometer marks while traveling at velocities of
2 m>s and then 10 m>s. Determine the train’s velocity when
it passes the next kilometer mark and the time it takes to
travel the 2-km distance.
Kinematics: For the first kilometer of the journey, v0 = 2 m>s, v = 10 m>s, s0 = 0,
and s = 1000 m. Thus,
+ B
A:
v2 = v0 2 + 2ac (s - s0)
102 = 22 + 2ac (1000 - 0)
ac = 0.048 m>s2
For the second
0.048 m>s2. Thus,
+ B
A:
v0 = 10 m>s,
kilometer,
s0 = 1000 m,
s = 2000 m,
and
v2 = v0 2 + 2ac (s - s0)
v2 = 102 + 2(0.048)(2000 - 1000)
v = 14 m>s
Ans.
For the whole journey, v0 = 2 m>s, v = 14 m>s, and 0.048 m>s2. Thus,
+ B
A:
v = v0 + act
14 = 2 + 0.048t
t = 250 s
Ans.
•12–17. A ball is thrown with an upward velocity of 5 m>s
from the top of a 10-m high building. One second later
another ball is thrown vertically from the ground with a
velocity of 10 m>s. Determine the height from the ground
where the two balls pass each other.
Kinematics: First, we will consider the motion of ball A with (vA)0 = 5 m>s,
(sA)0 = 0, sA = (h - 10) m, tA = t¿ , and ac = -9.81 m>s2. Thus,
A+cB
1
actA 2
2
sA = (sA)0 + (vA)0 tA +
h - 10 = 0 + 5t¿ +
1
(-9.81)(t¿)2
2
h = 5t¿ - 4.905(t¿)2 + 10
(1)
Motion of ball B is with (vB)0 = 10 m>s, (sB)0 = 0, sB = h, tB = t¿ - 1 and
ac = -9.81 m>s2. Thus,
A+cB
sB = (sB)0 + (vB)0 tB +
h = 0 + 10(t¿ - 1) +
1
ac tB 2
2
1
(-9.81)(t¿ - 1)2
2
h = 19.81t¿ - 4.905(t¿)2 - 14.905
(2)
Solving Eqs. (1) and (2) yields
h = 4.54 m
Ans.
t¿ = 1.68 m
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12–18. A car starts from rest and moves with a constant
acceleration of 1.5 m>s2 until it achieves a velocity of 25 m>s.
It then travels with constant velocity for 60 seconds.
Determine the average speed and the total distance traveled.
Kinematics: For stage (1) of the motion, v0 = 0, s0 = 0, v = 25 m>s, and
ac = 1.5 m>s2.
+ B
A:
v = v0 + act
25 = 0 + 1.5t1
t1 = 16.67 s
+ B
A:
v2 = v0 2 + 2ac(s - s0)
252 = 0 + 2(1.5)(s1 - 0)
s1 = 208.33 m
For stage (2) of the motion, s0 = 108.22 ft, v0 = 25 ft>s, t = 60 s, and ac = 0. Thus,
+ B
A:
s = s0 + v0t +
1
a t2
2 c
s = 208.33 + 25(60) + 0
= 1708.33ft = 1708 m
Ans.
The average speed of the car is then
vavg =
1708.33
s
=
= 22.3 m>s
t1 + t2
16.67 + 60
Ans.
12–19. A car is to be hoisted by elevator to the fourth
floor of a parking garage, which is 48 ft above the ground. If
the elevator can accelerate at 0.6 ft>s2, decelerate at
0.3 ft>s2, and reach a maximum speed of 8 ft>s, determine
the shortest time to make the lift, starting from rest and
ending at rest.
+c
v2 = v20 + 2 ac (s - s0)
v2max = 0 + 2(0.6)(y - 0)
0 = v2max + 2(-0.3)(48 - y)
0 = 1.2 y - 0.6(48 - y)
y = 16.0 ft,
+c
vmax = 4.382 ft>s 6 8 ft>s
v = v0 + ac t
4.382 = 0 + 0.6 t1
t1 = 7.303 s
0 = 4.382 - 0.3 t2
t2 = 14.61 s
t = t1 + t2 = 21.9 s
Ans.
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*12–20. A particle is moving along a straight line such that
its speed is defined as v = (-4s2) m>s, where s is in meters.
If s = 2 m when t = 0, determine the velocity and
acceleration as functions of time.
v = -4s2
ds
= -4s2
dt
L2
s
s - 2 ds =
L0
t
-4 dt
-s - 1| s2 = -4t|t0
t =
1 -1
(s - 0.5)
4
s =
2
8t + 1
v = -4 a
a =
2
2
16
b m>s
b = a8t + 1
(8t + 1)2
Ans.
16(2)(8t + 1)(8)
256
dv
= a
b m>s2
=
dt
(8t + 1)4
(8t + 1)3
Ans.
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•12–21. Two particles A and B start from rest at the origin
s = 0 and move along a straight line such that
aA = (6t - 3) ft>s2 and aB = (12t2 - 8) ft>s2, where t is in
seconds. Determine the distance between them when
t = 4 s and the total distance each has traveled in t = 4 s.
Velocity: The velocity of particles A and B can be determined using Eq. 12-2.
dyA = aAdt
L0
yA
dyA =
L0
t
(6t - 3)dt
yA = 3t2 - 3t
dyB = aBdt
L0
yB
dyB =
L0
t
(12t2 - 8)dt
yB = 4t3 - 8t
The times when particle A stops are
3t2 - 3t = 0
t = 0 s and = 1 s
The times when particle B stops are
t = 0 s and t = 22 s
4t3 - 8t = 0
Position: The position of particles A and B can be determined using Eq. 12-1.
dsA = yAdt
L0
sA
dsA =
L0
t
(3t2 - 3t)dt
sA = t3 -
3 2
t
2
dsB = yBdt
L0
sB
dsB =
L0
t
(4t3 - 8t)dt
sB = t4 - 4t2
The positions of particle A at t = 1 s and 4 s are
sA |t = 1 s = 13 -
3 2
(1 ) = -0.500 ft
2
sA |t = 4 s = 43 -
3 2
(4 ) = 40.0 ft
2
Particle A has traveled
dA = 2(0.5) + 40.0 = 41.0 ft
Ans.
The positions of particle B at t = 22 s and 4 s are
sB |t = 12 = (22)4 - 4(22)2 = -4 ft
sB |t = 4 = (4)4 - 4(4)2 = 192 ft
Particle B has traveled
dB = 2(4) + 192 = 200 ft
Ans.
At t = 4 s the distance beween A and B is
¢sAB = 192 - 40 = 152 ft
Ans.
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12–22. A particle moving along a straight line is subjected
to a deceleration a = (-2v3) m>s2, where v is in m>s. If it
has a velocity v = 8 m>s and a position s = 10 m when
t = 0, determine its velocity and position when t = 4 s.
Velocity: The velocity of the particle can be related to its position by applying
Eq. 12–3.
ds =
s
ydy
a
y
L10m
dy
- 2
L8m>s 2y
s - 10 =
1
1
2y
16
ds =
y =
8
16s - 159
[1]
Position: The position of the particle can be related to the time by applying Eq. 12–1.
dt =
L0
t
ds
y
s
dt =
L10m 8
1
(16s - 159) ds
8t = 8s2 - 159s + 790
When t = 4 s,
8(4) = 8s2 - 159s + 790
8s2 - 159s + 758 = 0
Choose the root greater than 10 m s = 11.94 m = 11.9 m
Ans.
Substitute s = 11.94 m into Eq. [1] yields
y =
8
= 0.250 m>s
16(11.94) - 159
Ans.
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12–23. A particle is moving along a straight line such that
its acceleration is defined as a = (-2v) m>s2, where v is in
meters per second. If v = 20 m>s when s = 0 and t = 0,
determine the particle’s position, velocity, and acceleration
as functions of time.
a = -2v
dv
= -2v
dt
v
t
dv
-2 dt
=
L20 v
L0
ln
v
= -2t
20
a =
L0
s
ds =
v = (20e - 2t)m>s
Ans.
dv
= (-40e - 2t)m>s2
dt
Ans.
L0
t
v dt =
L0
t
(20e - 2t)dt
s = -10e - 2t|t0 = -10(e - 2t - 1)
s = 10(1 - e - 2t)m
Ans.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–24. A particle starts from rest and travels along a
straight line with an acceleration a = (30 - 0.2v) ft>s2,
where v is in ft>s. Determine the time when the velocity of
the particle is v = 30 ft>s.
Velocity:
+ B
A:
dv
a
dt =
L0
v
t
dt =
dv
L0 30 - 0.2v
v
t|t0 = -
1
ln(30 - 0.2v) 2
0.2
0
t = 5ln
30
30 - 0.2v
t = 5ln
30
= 1.12 s
30 - 0.2(50)
Ans.
•12–25. When a particle is projected vertically upwards
with an initial velocity of v0, it experiences an acceleration
a = -(g + kv2) , where g is the acceleration due to gravity,
k is a constant and v is the velocity of the particle.
Determine the maximum height reached by the particle.
Position:
A+cB
ds =
L0
v dv
a
v
s
ds =
s|s0 = - c
s =
Lv0
-
vdv
g + kv2
v
1
ln A g + kv2 B d 2
2k
v0
g + kv0 2
1
ln ¢
≤
2k
g + kv2
The particle achieves its maximum height when v = 0. Thus,
hmax =
g + kv0 2
1
ln ¢
≤
g
2k
=
1
k
ln ¢ 1 + v0 2 ≤
g
2k
Ans.
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12–26. The acceleration of a particle traveling along a
straight line is a = (0.02et) m>s2, where t is in seconds. If
v = 0, s = 0 when t = 0, determine the velocity and
acceleration of the particle at s = 4 m.
Velocity:
A
+
:
a = 0.02e5.329 = 4.13 m>s2
B
Ans.
dv = a dt
L0
v
dv =
L0
t
0.02et dt
t
v|0 = 0.02e 2
v
t
0
v = C 0.02 A et - 1 B D m>s
(1)
Position:
+ B
A:
ds = v dt
L0
s
ds =
L0
t
0.02 A et - 1 B dt
s|0 = 0.02 A e - t B 2
s
t
t
0
s = 0.02 A et - t - 1 B m
When s = 4 m,
4 = 0.02 A et - t - 1 B
et - t - 201 = 0
Solving the above equation by trial and error,
t = 5.329 s
Thus, the velocity and acceleration when s = 4 m (t = 5.329 s) are
v = 0.02 A e5.329 - 1 B = 4.11 m>s
Ans.
a = 0.02e5.329 = 4.13 m>s2
Ans.
12–27. A particle moves along a straight line with an
acceleration of a = 5>(3s1>3 + s5>2) m>s2, where s is in
meters. Determine the particle’s velocity when s = 2 m, if it
starts from rest when s = 1 m. Use Simpson’s rule to
evaluate the integral.
a =
5
A 3s + s2 B
5
1
3
a ds = v dv
2
L1 A 3s + s
5 ds
1
3
0.8351 =
5
2
B
=
L0
v
v dv
1 2
v
2
v = 1.29 m>s
Ans.
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*12–28. If the effects of atmospheric resistance are
accounted for, a falling body has an acceleration defined by
the equation a = 9.81[1 - v2(10 -4)] m>s2, where v is in m>s
and the positive direction is downward. If the body is
released from rest at a very high altitude, determine (a) the
velocity when t = 5 s, and (b) the body’s terminal or
maximum attainable velocity (as t : q ).
Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2.
(+ T)
dt =
L0
t
dy
a
y
dy
2
L0 9.81[1 - (0.01y) ]
dt =
y
t =
y
dy
dy
1
c
+
d
9.81 L0 2(1 + 0.01y)
2(1
0.01y)
L0
9.81t = 50lna
y =
1 + 0.01y
b
1 - 0.01y
100(e0.1962t - 1)
[1]
e0.1962t + 1
a) When t = 5 s, then, from Eq. [1]
y =
b) If t : q ,
e0.1962t - 1
e0.1962t + 1
100[e0.1962(5) - 1]
e0.1962(5) + 1
= 45.5 m>s
Ans.
: 1. Then, from Eq. [1]
ymax = 100 m>s
Ans.
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•12–29. The position of a particle along a straight line is
given by s = (1.5t3 - 13.5t2 + 22.5t) ft, where t is in
seconds. Determine the position of the particle when
t = 6 s and the total distance it travels during the 6-s time
interval. Hint: Plot the path to determine the total distance
traveled.
Position: The position of the particle when t = 6 s is
s|t = 6s = 1.5(63) - 13.5(62) + 22.5(6) = -27.0 ft
Ans.
Total Distance Traveled: The velocity of the particle can be determined by applying
Eq. 12–1.
y =
ds
= 4.50t2 - 27.0t + 22.5
dt
The times when the particle stops are
4.50t2 - 27.0t + 22.5 = 0
t = 1s
and
t = 5s
The position of the particle at t = 0 s, 1 s and 5 s are
s冷t = 0s = 1.5(03) - 13.5(02) + 22.5(0) = 0
s冷t = 1s = 1.5(13) - 13.5(12) + 22.5(1) = 10.5 ft
s冷t = 5s = 1.5(53) - 13.5(52) + 22.5(5) = -37.5 ft
From the particle’s path, the total distance is
stot = 10.5 + 48.0 + 10.5 = 69.0 ft
Ans.
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12–30. The velocity of a particle traveling along a straight
line is v = v0 - ks, where k is constant. If s = 0 when t = 0,
determine the position and acceleration of the particle as a
function of time.
Position:
+ B
A:
ds
y
dt =
L0
t
s
dt =
t冷t0 = t =
ds
L0 v0 - ks
s
1
ln (v0 - ks) 2
k
0
v0
1
ln ¢
≤
k
v0 - ks
ekt =
v0
v0 - ks
s =
v0
A 1 - e - kt B
k
v =
ds
d v0
=
c A 1 - e - kt B d
dt
dt k
Ans.
Velocity:
v = v0e - kt
Acceleration:
a =
d
dv
=
A v e - kt B
dt
dt 0
a = -kv0e - kt
Ans.
12–31. The acceleration of a particle as it moves along a
straight line is given by a = 12t - 12 m>s2, where t is in
seconds. If s = 1 m and v = 2 m>s when t = 0, determine the
particle’s velocity and position when t = 6 s. Also, determine
the total distance the particle travels during this time period.
L2
v
dv =
L0
t
(2 t - 1) dt
v = t2 - t + 2
L1
s
ds =
s =
L0
t
(t2 - t + 2) dt
1 3
1
t - t2 + 2 t + 1
3
2
When t = 6 s,
v = 32 m>s
Ans.
s = 67 m
Ans.
d = 67 - 1 = 66 m
Ans.
Since v Z 0 then
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*12–32. Ball A is thrown vertically upward from the top
of a 30-m-high-building with an initial velocity of 5 m>s. At
the same instant another ball B is thrown upward from the
ground with an initial velocity of 20 m>s. Determine the
height from the ground and the time at which they pass.
Origin at roof:
Ball A:
A+cB
s = s0 + v0 t +
-s = 0 + 5t -
1
a t2
2 c
1
(9.81)t2
2
Ball B:
A+cB
s = s0 + v0 t +
1
a t2
2 c
-s = -30 + 20t -
1
(9.81)t2
2
Solving,
t = 2s
Ans.
s = 9.62 m
Distance from ground,
d = (30 - 9.62) = 20.4 m
Ans.
Also, origin at ground,
s = s0 + v0 t +
1
a t2
2 c
sA = 30 + 5t +
1
(-9.81)t2
2
sB = 0 + 20t +
1
(-9.81)t2
2
Require
sA = sB
30 + 5t +
1
1
(-9.81)t2 = 20t + (-9.81)t2
2
2
t = 2s
Ans.
sB = 20.4 m
Ans.
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•12–33. A motorcycle starts from rest at t = 0 and travels
along a straight road with a constant acceleration of 6 ft>s2
until it reaches a speed of 50 ft>s. Afterwards it maintains
this speed. Also, when t = 0, a car located 6000 ft down the
road is traveling toward the motorcycle at a constant speed
of 30 ft>s. Determine the time and the distance traveled by
the motorcycle when they pass each other.
Motorcycle:
+ B
A:
v = v0 + ac t¿
50 = 0 + 6t¿
t¿ = 8.33 s
v2 = v20 + 2ac (s - s0)
(50)2 = 0 + 2(6)(s¿ - 0)
s¿ = 208.33 ft
In t¿ = 8.33 s car travels
s– = v0 t¿ = 30(8.33) = 250 ft
Distance between motorcycle and car:
6000 - 250 - 208.33 = 5541.67 ft
When passing occurs for motorcycle,
s = v0 t;
x = 50(t–)
For car:
s = v0 t;
5541.67 - x = 30(t–)
Solving,
x = 3463.54 ft
t– = 69.27 s
Thus, for the motorcycle,
t = 69.27 + 8.33 = 77.6 s
Ans.
sm = 208.33 + 3463.54 = 3.67(10)3 ft
Ans.
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12–34. A particle moves along a straight line with a
velocity v = (200s) mm>s, where s is in millimeters.
Determine the acceleration of the particle at s = 2000 mm.
How long does the particle take to reach this position if
s = 500 mm when t = 0?
Acceleration:
+ B
A:
dv
= 200s
ds
Thus, a = v
dv
= (200s)(200) = 40 A 103 B s mm>s2
ds
When s = 2000 mm,
a = 40 A 103 B (2000) = 80 A 106 B mm>s2 = 80 km>s2
Ans.
Position:
+ B
A:
ds
v
dt =
t
L0
s
dt =
t
ds
L500 mm 200s
s
t2 =
0
t =
1
lns 2
200
500 mm
1
s
ln
200 500
At s = 2000 mm,
t =
2000
1
ln
= 6.93 A 10 - 3 B s = 6.93 ms
200
500
Ans.
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A particle has an initial speed of 27 m>s. If it
experiences a deceleration of a = 1-6t2 m>s2, where t is in
seconds, determine its velocity, after it has traveled 10 m.
How much time does this take?
쐍12–35.
Velocity:
+ B
A:
dv = a dt
t
L27
v2
dv =
v
=
27
L0
t
(-6t)dt
A -3t2 B 2
t
0
2
v = (27 - 3t ) m>s
+ B
A:
ds = v dt
L0
s
ds =
L0
t
A 27 - 3t2 B dt
s 2 = A 27t - t3 B 2
s
0
t
0
3
s = (27t - t ) m>s
When s = 100 m,
t = 0.372 s
Ans.
v = 26.6 m>s
Ans.
*12–36. The acceleration of a particle traveling along a
straight line is a = (8 - 2s) m>s2, where s is in meters. If
v = 0 at s = 0, determine the velocity of the particle at
s = 2 m, and the position of the particle when the velocity
is maximum.
Velocity:
+ B
A:
v dv = a ds
L0
v
vdv =
L0
s
(8 - 2s) ds
v
s
v2
` = A 8s - s2 B 2
2 0
0
n = 216s - 2s2 m>s
At s = 2 m,
v冷s = 2 m = 216(2) - 2 A 22 B = ;4.90 m>s
When the velocity is maximum
Ans.
dv
= 0. Thus,
ds
dv
16 - 4s
=
= 0
ds
2 216s - 2s2
16 - 4s = 0
s = 4m
Ans.
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•12–37. Ball A is thrown vertically upwards with a
velocity of v0. Ball B is thrown upwards from the same point
with the same velocity t seconds later. Determine the
elapsed time t 6 2v0>g from the instant ball A is thrown to
when the balls pass each other, and find the velocity of each
ball at this instant.
Kinematics: First, we will consider the motion of ball A with (vA)0 = v0, (sA)0 = 0,
sA = h, tA = t¿ , and (ac)A = -g.
A+cB
h = 0 + v0t¿ +
h = v0t¿ -
A+cB
1
(a ) t 2
2 cA A
sA = (sA)0 + (vA)0tA +
1
(-g)(t¿)2
2
g 2
t¿
2
(1)
vA = (vA)0 + (ac)A tA
vA = v0 + (-g)(t¿)
vA = v0 - gt¿
(2)
The motion of ball B requires (vB)0 = v0, (sB)0 = 0, sB = h, tB = t¿ - t , and
(ac)B = -g.
A+cB
sB = (sB)0 + (vB)0tB +
h = 0 + v0(t¿ - t) +
h = v0(t¿ - t) -
A+cB
1
(a ) t 2
2 cBB
1
(-g)(t¿ - t)2
2
g
(t¿ - t)2
2
(3)
vB = (vB)0 + (ac)B tB
vB = v0 + (-g)(t¿ - t)
vB = v0 - g(t¿ - t)
(4)
Solving Eqs. (1) and (3),
g 2
g
t¿ = v0(t¿ - t) - (t¿ - t)2
2
2
2v0 + gt
t¿ =
2g
v0t¿ -
Ans.
Substituting this result into Eqs. (2) and (4),
vA = v0 - ga
= -
1
1
gt = gt T
2
2
vB = v0 - ga
=
2v0 + gt
b
2g
Ans.
2v0 + gt
- tb
2g
1
gt c
2
Ans.
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12–38. As a body is projected to a high altitude above the
earth’s surface, the variation of the acceleration of gravity with
respect to altitude y must be taken into account. Neglecting air
resistance, this acceleration is determined from the formula
a = -g0[R2>(R + y)2], where g0 is the constant gravitational
acceleration at sea level, R is the radius of the earth, and the
positive direction is measured upward. If g0 = 9.81 m>s2 and
R = 6356 km, determine the minimum initial velocity (escape
velocity) at which a projectile should be shot vertically from
the earth’s surface so that it does not fall back to the earth.
Hint: This requires that v = 0 as y : q .
v dv = a dy
Ly
q
0
2
v dv = -g0R
dy
L0 (R + y)
2
g0 R2 q
v2 2 0
2
=
2 y
R + y 0
v = 22g0 R
= 22(9.81)(6356)(10)3
= 11167 m>s = 11.2 km>s
Ans.
12–39. Accounting for the variation of gravitational
acceleration a with respect to altitude y (see Prob. 12–38),
derive an equation that relates the velocity of a freely falling
particle to its altitude.Assume that the particle is released from
rest at an altitude y0 from the earth’s surface.With what velocity
does the particle strike the earth if it is released from rest at an
altitude y0 = 500 km? Use the numerical data in Prob. 12–38.
From Prob. 12–38,
(+ c )
a = -g0
R2
(R + y)2
Since a dy = v dv
then
y
-g0 R2
dy
2
Ly0 (R + y)
=
L0
v
v dv
g0 R2 c
y
v2
1
d =
R + y y0
2
g0 R2[
1
1
v2
] =
R + y
R + y0
2
Thus
2g0 (y0 - y)
v = -R
A (R + y)(R + y0)
When y0 = 500 km,
y = 0,
2(9.81)(500)(103)
v = -6356(103)
A 6356(6356 + 500)(106)
v = -3016 m>s = 3.02 km>s T
Ans.
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*12–40. When a particle falls through the air, its initial
acceleration a = g diminishes until it is zero, and thereafter it
falls at a constant or terminal velocity vf. If this variation of
the acceleration can be expressed as a = 1g>v2f21v2f - v22,
determine the time needed for the velocity to become
v = vf>2 . Initially the particle falls from rest.
g
dv
= a = ¢ 2 ≤ A v2f - v2 B
dt
vf
v
L0 v2f - v2¿
dy
=
g
v2f
L0
t
dt
vf + v y
g
1
ln ¢
≤` = 2t
2vf
vf - v 0
vf
t =
t =
vf
2g
vf
2g
ln ¢
ln ¢
t = 0.549a
vf + v
vf - v
≤
vf + vf>2
vf - vf>2
vf
g
≤
b
Ans.
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•12–41. A particle is moving along a straight line such that
its position from a fixed point is s = (12 - 15t2 + 5t3) m,
where t is in seconds. Determine the total distance traveled
by the particle from t = 1 s to t = 3 s. Also, find the average
speed of the particle during this time interval.
Velocity:
+ B
A:
v =
ds
d
=
A 12 - 15t2 + 5t3 B
dt
dt
v = -30t + 15t2 m>s
The velocity of the particle changes direction at the instant when it is momentarily
brought to rest. Thus,
v = -30t + 15t2 = 0
t(-30 + 15t) = 0
t = 0 and 2 s
Position: The positions of the particle at t = 0 s, 1 s, 2 s, and 3 s are
s冷t = 0 s = 12 - 15 A 02 B + 5 A 03 B = 12 m
s冷t = 1 s = 12 - 15 A 12 B + 5 A 13 B = 2 m
s冷t = 2 s = 12 - 15 A 22 B + 5 A 23 B = -8 m
s冷t = 3 s = 12 - 15 A 32 B + 5 A 33 B = 12 m
Using the above results, the path of the particle is shown in Fig. a. From this figure,
the distance traveled by the particle during the time interval t = 1 s to t = 3 s is
sTot = (2 + 8) + (8 + 12) = 30 m
Ans.
The average speed of the particle during the same time interval is
vavg =
sTot
30
=
= 15 m>s
¢t
3 - 1
Ans.
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12–42. The speed of a train during the first minute has
been recorded as follows:
t (s)
0
20
40
60
v (m>s)
0
16
21
24
Plot the v-t graph, approximating the curve as straight-line
segments between the given points. Determine the total
distance traveled.
The total distance traveled is equal to the area under the graph.
sT =
1
1
1
(20)(16) + (40 - 20)(16 + 21) + (60 - 40)(21 + 24) = 980 m
2
2
2
28
Ans.
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12–43. A two-stage missile is fired vertically from rest
with the acceleration shown. In 15 s the first stage A burns
out and the second stage B ignites. Plot the v-t and s-t
graphs which describe the two-stage motion of the missile
for 0 … t … 20 s.
B
A
a (m/s2)
25
18
t (s)
15
a dt, the constant lines of the a–t graph become sloping lines for the v–t graph.
L
The numerical values for each point are calculated from the total area under the a–t graph to
the point.
Since v =
At t = 15 s,
v = (18)(15) = 270 m>s
At t = 20 s,
v = 270 + (25)(20 - 15) = 395 m>s
v dt, the sloping lines of the v–t graph become parabolic curves for the s–t graph.
L
The numerical values for each point are calculated from the total area under the v–t graph to
the point.
Since s =
1
(15)(270) = 2025 m
2
At t = 15 s,
s =
At t = 20 s,
s = 2025 + 270(20 - 15) +
1
(395 - 270)(20 - 15) = 3687.5 m = 3.69 km
2
Also:
0 … t … 15:
a = 18
v = v0 + ac t = 0 + 18t
s = s0 + v0 t +
1
a t2 = 0 + 0 + 9t2
2 c
At t = 15:
v = 18(15) = 270
s = 9(15)2 = 2025
15 … t … 20:
a = 25
v = v0 + ac t = 270 + 25(t - 15)
s = s0 + v0 t +
1
1
a t2 = 2025 + 270(t - 15) + (25)(t - 15)2
2 c
2
When t = 20:
v = 395 m>s
s = 3687.5 m = 3.69 km
29
20
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*12–44. A freight train starts from rest and travels with a
constant acceleration of 0.5 ft>s2. After a time t¿ it maintains a
constant speed so that when t = 160 s it has traveled 2000 ft.
Determine the time t¿ and draw the v–t graph for the motion.
Total Distance Traveled: The distance for part one of the motion can be related to
time t = t¿ by applying Eq. 12–5 with s0 = 0 and y0 = 0.
+ B
A:
s = s0 + y0 t +
s1 = 0 + 0 +
1
a t2
2 c
1
(0.5)(t¿)2 = 0.25(t¿)2
2
The velocity at time t can be obtained by applying Eq. 12–4 with y0 = 0.
+ B
A:
y = y0 + act = 0 + 0.5t = 0.5t
[1]
The time for the second stage of motion is t2 = 160 - t¿ and the train is traveling at
a constant velocity of y = 0.5t¿ (Eq. [1]). Thus, the distance for this part of motion is
+ B
A:
s2 = yt2 = 0.5t¿(160 - t¿) = 80t¿ - 0.5(t¿)2
If the total distance traveled is sTot = 2000, then
sTot = s1 + s2
2000 = 0.25(t¿)2 + 80t¿ - 0.5(t¿)2
0.25(t¿)2 - 80t¿ + 2000 = 0
Choose a root that is less than 160 s, then
t¿ = 27.34 s = 27.3 s
Ans.
Yⴚt Graph: The equation for the velocity is given by Eq. [1]. When t = t¿ = 27.34 s,
y = 0.5(27.34) = 13.7 ft>s.
30
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•12–45. If the position of a particle is defined by
s = [2 sin (p>5)t + 4] m, where t is in seconds, construct
the s-t, v-t, and a-t graphs for 0 … t … 10 s.
31
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12–46. A train starts from station A and for the first
kilometer, it travels with a uniform acceleration. Then, for
the next two kilometers, it travels with a uniform speed.
Finally, the train decelerates uniformly for another
kilometer before coming to rest at station B. If the time for
the whole journey is six minutes, draw the v–t graph and
determine the maximum speed of the train.
For stage (1) motion,
+ B
A:
v1 = v0 + (ac)1 t
vmax = 0 + (ac)1 t1
vmax = (ac)1t1
+ B
A:
(1)
v1 2 = v0 2 + 2(ac)1(s1 - s0)
vmax 2 = 0 + 2(ac)1(1000 - 0)
(ac)1 =
vmax 2
2000
(2)
Eliminating (ac)1 from Eqs. (1) and (2), we have
t1 =
2000
vmax
(3)
For stage (2) motion, the train travels with the constant velocity of vmax for
t = (t2 - t1). Thus,
+ B
A:
1
(a ) t2
2 c2
s2 = s1 + v1t +
1000 + 2000 = 1000 + vmax (t2 - t1) + 0
t2 - t1 =
2000
vmax
(4)
For stage (3) motion, the train travels for t = 360 - t2. Thus,
+ B
A:
v3 = v2 + (ac)3t
0 = vmax - (ac)3(360 - t2)
vmax = (ac)3(360 - t2)
+ B
A:
(5)
v3 2 = v2 2 + 2(ac)3(s3 - s2)
0 = vmax 2 + 2 C -(ac)3 D (4000 - 3000)
(ac)3 =
vmax 2
2000
(6)
Eliminating (ac)3 from Eqs. (5) and (6) yields
360 - t2 =
2000
vmax
(7)
Solving Eqs. (3), (4), and (7), we have
t1 = 120 s
t2 = 240 s
vmax = 16.7 m>s
Ans.
Based on the above results the v-t graph is shown in Fig. a.
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12–47. The particle travels along a straight line with the
velocity described by the graph. Construct the a-s graph.
v (m/s)
13
10
4
a = v
3
dv
= (2s + 4)(2) = (4s + 8) m>s2
ds
At s = 0 m and 3 m,
a|s = 0 m = 4(0) + 8 = 8 m>s2
a|s = 3 m = 4(3) + 8 = 20 m>s2
For 3m 6 s … 6 m,
+ B
A:
a = v
v ⫽ 2s ⫹ 4
s (m)
aⴚs Graph: For 0 … s 6 3 m,
+ B
A:
v⫽s⫹7
dv
= (s + 7)(1) = (s + 7) m>s2
ds
At s = 3 m and 6 m,
a|s = 3 m = 3 + 7 = 10 m>s2
a|s = 6 m = 6 + 7 = 13 m>s2
The a-s graph is shown in Fig. a.
33
6
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*12–48. The a–s graph for a jeep traveling along a straight
road is given for the first 300 m of its motion. Construct the
v–s graph. At s = 0, v = 0.
a (m/s2)
2
200
aⴚs Graph: The function of acceleration a in terms of s for the interval
0 m … s 6 200 m is
a - 0
2 - 0
=
s - 0
200 - 0
a = (0.01s) m>s2
For the interval 200 m 6 s … 300 m,
a - 2
0 - 2
=
s - 200
300 - 200
a = (-0.02s + 6) m>s2
Yⴚs Graph: The function of velocity y in terms of s can be obtained by applying
ydy = ads. For the interval 0 m ◊ s<200 m,
ydy = ds
L0
y
ydy =
L0
s
0.01sds
y = (0.1s) m>s
y = 0.100(200) = 20.0 m>s
At s = 200 m,
For the interval 200 m 6 s … 300 m,
ydy = ads
s
y
L20.0m>s
y =
At s = 300 m,
ydy =
L200m
(-0.02s + 6)ds
A 2 -0.02s2 + 12s - 1200 B m>s
y = 2 -0.02(3002) + 12(300) - 1200 = 24.5 m>s
34
300
s (m)
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•12–49. A particle travels along a curve defined by the
equation s = (t3 - 3t2 + 2t) m. where t is in seconds. Draw
the s - t, v - t, and a - t graphs for the particle for
0 … t … 3 s.
s = t3 - 3t2 + 2t
v =
ds
= 3t2 - 6t + 2
dt
a =
dv
= 6t - 6
dt
v = 0 at 0 = 3t2 - 6t + 2
t = 1.577 s, and t = 0.4226 s,
s|t = 1.577 = -0.386 m
s|t = 0.4226 = 0.385 m
12–50. A truck is traveling along the straight line with a
velocity described by the graph. Construct the a-s graph
for 0 … s … 1500 ft.
v (ft/s)
v ⫽ 0.6 s3/4
aⴚs Graph: For 0 … s 6 625 ft,
+ B
A:
a = v
75
dv
3
= A 0.6s3>4 B c (0.6)s - 1>4 d = A 0.27s1>2 B ft >s2
ds
4
At s = 625ft,
a|s = 625 ft = 0.27 A 6251>2 B = 6.75ft>s2
For 625 ft 6 s 6 1500 ft,
+ B
A:
s(ft)
625
a = v
dv
= 75(0) = 0
ds
The a-s graph is shown in Fig. a.
35
1500
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12–51. A car starts from rest and travels along a straight
road with a velocity described by the graph. Determine the
total distance traveled until the car stops. Construct the s–t
and a–t graphs.
v(m/s)
30
v ⫽ ⫺0.5t ⫹ 45
v⫽t
sⴚt Graph: For the time interval 0 … t 6 30 s, the initial condition is s = 0 when
t = 0 s.
+ B
A:
ds = vdt
L0
s
ds =
L0
t
tdt
t2
s = ¢ ≤m
2
When t = 30 s,
s =
302
= 450 m
2
For the time interval 30 s 6 t … 90 s, the initial condition is s = 450 m when
t = 30 s.
+ B
A:
ds = vdt
s
t
(-0.5t + 45)dt
L30s
L450 m
ds =
1
s = a - t2 + 45t - 675b m
4
When t = 90 s,
s冷t = 90 s = -
1
A 902 B + 45(90) - 675 = 1350 m
4
Ans.
The s - t graph shown is in Fig. a.
aⴚt Graph: For the time interval 0 6 t 6 30 s,
a =
dv
d
=
(t) = 1 m>s2
dt
dt
For the time interval 30 s 6 t … 90 s,
a =
dv
d
=
(-0.5t + 45) = -0.5 m>s2
dt
dt
The a- t graph is shown in Fig. b.
Note: Since the change in position of the car is equal to the area under the v - t
graph, the total distance traveled by the car is
¢s =
L
vdt
s冷t = 90 s - 0 =
1
(90)(30)
2
s冷t = 90 s = 1350 s
36
t(s)
30
60
90
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*12–52. A car travels up a hill with the speed shown.
Determine the total distance the car travels until it stops
(t = 60 s). Plot the a-t graph.
v (m/s)
10
Distance traveled is area under v - t graph.
s = (10)(30) +
30
1
(10)(30) = 450 m
2
60
t (s)
Ans.
•12–53. The snowmobile moves along a straight course
according to the v–t graph. Construct the s–t and a–t graphs
for the same 50-s time interval. When t = 0, s = 0.
v (m/s)
sⴚt Graph: The position function in terms of time t can be obtained by applying
ds
2
12
y =
. For time interval 0 s … t 6 30 s, y =
t = a t b m>s.
dt
30
5
12
ds = ydt
L0
t
s
ds =
2
tdt
L0 5
30
1
s = a t2 b m
5
At t = 30 s ,
s =
1
A 302 B = 180 m
5
For time interval 30 s<t ◊ 50 s,
ds = ydt
s
L180 m
ds =
t
12dt
L30 s
s = (12t - 180) m
At t = 50 s,
s = 12(50) - 180 = 420 m
aⴚt Graph: The acceleration function in terms of time t can be obtained by applying
dy
dy
2
a =
. For time interval 0 s ◊ t<30 s and 30 s<t ◊ 50 s, a =
=
dt
dt
5
dy
= 0.4 m>s2 and a =
= 0, respectively.
dt
37
50
t (s)
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12–54. A motorcyclist at A is traveling at 60 ft>s when he
wishes to pass the truck T which is traveling at a constant
speed of 60 ft>s. To do so the motorcyclist accelerates at
6 ft>s2 until reaching a maximum speed of 85 ft>s. If he then
maintains this speed, determine the time needed for him to
reach a point located 100 ft in front of the truck. Draw the
v -t and s-t graphs for the motorcycle during this time.
(vm)1 ⫽ 60 ft/s
(vm)2 ⫽ 85 ft/s
vt ⫽ 60 ft/s
A
T
40 ft
Motorcycle:
Time to reach 85 ft>s,
v = v0 + ac t
85 = 60 + 6t
t = 4.167 s
v2 = v20 + 2ac (s - s0)
Distance traveled,
(85)2 = (60)2 + 2(6)(sm - 0)
sm = 302.08 ft
In t = 4.167 s, truck travels
st = 60(4.167) = 250 ft
Further distance for motorcycle to travel: 40 + 55 + 250 + 100 - 302.08
= 142.92 ft
Motorcycle:
s = s0 + v0 t
(s + 142.92) = 0 + 85t¿
Truck:
s = 0 + 60t¿
Thus t¿ = 5.717 s
t = 4.167 + 5.717 = 9.88 s
Ans.
Total distance motorcycle travels
sT = 302.08 + 85(5.717) = 788 ft
38
55 ft
100 ft
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12–55. An airplane traveling at 70 m>s lands on a straight
runway and has a deceleration described by the graph.
Determine the time t¿ and the distance traveled for it to
reach a speed of 5 m>s. Construct the v–t and s-t graphs for
this time interval, 0 … t … t¿ .
a(m/s2)
5
⫺4
Nⴚt Graph: For the time interval 0 … t 6 5 s, the initial condition is v = 70 m>s
when t = 0 s.
+ B
A:
dv = adt
v
L70 m>s
dv =
L0
t
-10dt
v = ( -10t + 70) m>s
When t = 5 s,
v|t = 5 s = -10(5) + 70 = 20 m>s
For the time interval 5 s 6 t … t¿ , the initial condition is v = 20 m>s when t = 5 s.
+ B
A:
dv = adt
v
L20 m>s
t
dv =
L5 s
-4dt
v = (-4t + 40) m>s
When v = 5 m>s,
5 = -4t¿ + 40
t¿ = 8.75 s
Ans.
Also, the change in velocity is equal to the area under the a–t graph. Thus,
¢v =
L
adt
5 - 70 = - C 5(10) + 4(t¿ - 5) D
t¿ = 8.75s¿
The v–t graph is shown in Fig. a.
39
⫺10
tÀ
t(s)
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12–57. Continued
sⴚt Graph: For the time interval 0 … t 6 5 s, the initial condition is s = 0 when
t = 0 s.
+ B
A:
ds = vdt
L0
s
ds =
s =
L0
t
(-10t + 70)dt
A -5t2 + 70t B m
When t = 5 s,
s冷t = 5 s = -5 A 52 B + 70(5) = 225 m
For the time interval 5 6 t … t¿ = 8.75 s the initial condition is s = 225 m when
t = 5 s.
+ B
A:
ds = vdt
s
L225 m
ds =
s =
L5
t
(-4t + 40)dt
A -2t2 + 40t + 75 B m
When t = t¿ = 8.75 s,
s冷t = 8.75 s = -2 A 8.752 B + 40(8.75) + 75 = 271.875 m = 272 m
Ans.
Also, the change in position is equal to the area under the v–t graph. Referring to
Fig. a, we have
¢s =
L
vdt
s冷t = 8.75 s - 0 =
1
1
(70 + 20)(5) + (20 + 5)(3.75) = 271.875 m = 272 m
2
2
The s–t graph is shown in Fig. b.
40
Ans.
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*12–56. The position of a cyclist traveling along a straight
road is described by the graph. Construct the v–t and a–t
graphs.
s (m)
137.5
s ⫽ ⫺0.625 t2 ⫹ 27.5t ⫺ 162.5
Nⴚt Graph: For the time interval 0 … t 6 10 s,
+ B
A:
v =
ds
d
=
A 0.05t3 B = A 0.15t2 B m>s
dt
dt
When t = 0 s and 10 s,
v冷t = 0 = 0.15 A 0
50
2
B =0
v冷t = 10 s = 0.15 A 10
2
B = 15 m/s
For the time interval 10 s 6 t … 20 s,
+ B
A:
v =
ds
d
=
A -0.625t2 + 27.5t - 162.5 B = (-1.25t + 27.5) m>s
dt
dt
When t = 10 s and 20 s,
v冷t = 10 s = -1.25(10) + 27.5 = 15 m>s
v冷t = 20 s = -1.25(20) + 27.5 = 2.5 m>s
The v–t graph is shown in Fig. a.
aⴚt Graph: For the time interval 0 … t 6 10 s,
+ B
A:
a =
dv
d
=
A 0.15t2 B = (0.3t) m>s2
dt
dt
When t = 0 s and 10 s,
a冷 t = 0 s = 0.3(0) = 0
a冷 t = 10 s = 0.3(10) = 3 m>s2
For the time interval 10 s 6 t … 20 s,
+ B
A:
a =
dv
d
=
(-1.25t + 27.5) = -1.25 m>s2
dt
dt
the a–t graph is shown in Fig. b.
41
s ⫽ 0.05 t3
t (s)
10
20
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•12–57. The dragster starts from rest and travels along a
straight track with an acceleration-deceleration described
by the graph. Construct the v-s graph for 0 … s … s¿, and
determine the distance s¿ traveled before the dragster again
comes to rest.
a(m/s2)
25
a ⫽ 0.1s ⫹ 5
5
N ⴚs Graph: For 0 … s 6 200 m, the initial condition is v = 0 at s = 0.
+ B
A:
200
vdv = ads
L0
v
vdv =
⫺15
L0
s
(0.1s + 5)ds
v
s
v2 2
= A 0.05s2 + 5s B 2
2 0
0
v = a 20.1s2 + 10sb m>s
At s = 200 m,
v冷s = 200 m = 20.1 A 2002 B + 10(200) = 77.46 m>s = 77.5 m>s
For 200 m 6 s … s¿ , the initial condition is v = 77.46 m>s at s = 200 m.
+ B
A:
vdv = ads
v
L77.46 m>s
s
v dv =
L200 m
-15ds
v
s
v2 2
= -15s冷200 m
2 77.46 m>s
v =
A 2-30s + 12000 B m>s
When v = 0,
0 = 2 -30s¿ + 12000
s¿ = 400 m
Ans.
The v–s graph is shown in Fig. a.
42
s
s (m)
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12–58. A sports car travels along a straight road with an
acceleration-deceleration described by the graph. If the car
starts from rest, determine the distance s¿ the car travels
until it stops. Construct the v -s graph for 0 … s … s¿ .
a(ft/s2)
6
1000
⫺4
Nⴚs Graph: For 0 … s 6 1000 ft, the initial condition is v = 0 at s = 0.
+ B
A:
vdv = ads
L0
v
vdv =
L0
s
6ds
v2
= 6s
2
A 212s1>2 B ft>s
v =
When s = 1000 ft,
v = 212(1000)1>2 = 109.54 ft>s = 110 ft>s
For 1000 ft 6 s … s¿ , the initial condition is v = 109.54 ft>s at s = 1000 ft.
+ B
A:
vdv = ads
s
v
L109.54 ft>s
vdv =
L1000 ft
-4ds
v
s
v2 2
= -4s冷1000 ft
2 109.54 ft>s
v =
A 220 000 - 8s B ft>s
When v = 0,
0 = 220 000 - 8s¿
s¿ = 2500 ft
Ans.
The v–s graph is shown in Fig. a.
43
s
s(ft)
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12–59. A missile starting from rest travels along a straight
track and for 10 s has an acceleration as shown. Draw the
v-t graph that describes the motion and find the distance
traveled in 10 s.
a (m/s2)
40
a ⫽ 2t ⫹ 20
30
For t … 5 s,
a ⫽ 6t
a = 6t
dv = a dt
L0
v
L0
dv =
t
6t dt
t(s)
5
v = 3t2
When t = 5 s,
v = 75 m>s
For 5 6 t 6 10 s,
a = 2t + 20
dv = a dt
v
L75
dv =
L5
t
(2t + 20) dt
v - 75 = t2 + 20t - 125
v = t2 + 20t - 50
When t = 10 s,
v = 250 m>s
Distance at t = 5 s:
ds = v dt
L0
s
ds =
L0
5
3t2 dt
s = (5)3 = 125 m
Distance at t = 10 s:
ds = v dv
s
L125
ds =
s - 125 =
L5
10
A t2 + 20t - 50 B dt
10
1 3
t + 10t2 - 50t d
3
5
s = 917 m
Ans.
44
10
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*12–60. A motorcyclist starting from rest travels along a
straight road and for 10 s has an acceleration as shown.
Draw the v-t graph that describes the motion and find the
distance traveled in 10 s.
For 0 … t 6 6
a (m/s2)
dv = a dt
v
6
t
1 2
dv =
t dt
L0 6
L0
v =
1 3
t
18
6
ds = v dt
L0
t
s
ds =
s =
When t = 6 s,
1 3
t dt
18
L0
1 4
t
72
v = 12 m>s
For 6 6 t … 10
s = 18 m
dv = a dt
v
L12
dv =
L6
t
6 dt
v = 6t - 24
ds = v dt
s
L18
ds =
L6
t
(6t - 24) dt
s = 3t2 - 24t + 54
When t = 10 s,
1 2
a ⫽—
t
6
v = 36 m>s
s = 114 m
Ans.
45
10
t (s)
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•12–61. The v-t graph of a car while traveling along a
road is shown. Draw the s-t and a-t graphs for the motion.
v (m/s)
20
0 … t … 5
¢v
20
=
= 4 m>s2
¢t
5
a =
5 … t … 20
a =
20 … t … 30
a =
5
20
20 - 20
¢v
=
= 0 m>s2
¢t
20 - 5
¢v
0 - 20
=
= -2 m>s2
¢t
30 - 20
From the v–t graph at t1 = 5 s, t2 = 20 s, and t3 = 30 s,
s1 = A1 =
1
(5)(20) = 50 m
2
s2 = A1 + A2 = 50 + 20(20 - 5) = 350 m
s3 = A1 + A2 + A3 = 350 +
1
(30 - 20)(20) = 450 m
2
The equations defining the portions of the s–t graph are
0 … t … 5s
v = 4t;
ds = v dt;
L0
s
ds =
L0
s
5 … t … 20 s
v = 20;
ds = v dt;
L50
ds =
t
s = 2t2
4t dt;
L5
t
t
s
20 … t … 30 s
v = 2(30 - t);
ds = v dt;
L350
s = 20t - 50
20 dt;
ds =
L20
46
2(30 - t) dt;
s = -t2 + 60t - 450
30
t (s)
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12–62. The boat travels in a straight line with the
acceleration described by the a-s graph. If it starts from rest,
construct the v-s graph and determine the boat’s maximum
speed. What distance s¿ does it travel before it stops?
a(m/s2)
6
a ⫽ ⫺0.02s ⫹ 6
3
Nⴚs Graph: For 0 … s 6 150 m, the initial condition is v = 0 at s = 0.
+ B
A:
vdv = ads
L0
v
vdv =
v
v2 2
=
2 0
150
L0
s
( -0.02s + 6)ds
⫺4
A -0.01s2 + 6s B 冷0
s
v = a 2-0.02s2 + 12sb m>s
The maximum velocity of the boat occurs at s = 150 m, where its acceleration
changes sign. Thus,
vmax = v冷s = 150 m = 2 -0.02 A 1502 B + 12(150) = 36.74 m>s = 36.7 m>s
Ans.
For 150 m 6 s 6 s¿ , the initial condition is v = 36.74 m>s at s = 150 m.
+ B
A:
vdv = ads
v
L36.74 m>s
vdv =
s
L150 m
v
-4ds
s
v2 2
= -4s 2
2 36.74 m>s
150 m
v = 2-8s + 2550 m>s
Thus, when v = 0,
0 = 2 -8s¿ + 2550
s¿ = 318.7 m = 319 m
The v–s graph is shown in Fig. a.
47
Ans.
s
s(m)
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12–63. The rocket has an acceleration described by the
graph. If it starts from rest, construct the v-t and s-t
graphs for the motion for the time interval 0 … t … 14 s.
a(m/s2)
Nⴚt Graph: For the time interval 0 … t 6 9 s, the initial condition is v = 0 at s = 0.
A+cB
38
dv = adt
L0
v
dv =
a2 ⫽ 36t
L0
a ⫽ 4t ⫺ 18
t
1>2
6t dt
18
v = A 4t3>2 B m>s
t(s)
When t = 9 s,
9
v冷t = 9 s = 4 A 93>2 B = 108 m>s
The initial condition is v = 108 m>s at t = 9 s.
A+cB
dv = adt
v
t
L108 m>s
L9 s
dv =
(4t - 18)dt
v = A 2t2 - 18t + 108 B m>s
When t = 14 s,
v冷t = 14 s = 2 A 142 B - 18(14) + 108 = 248 m>s
The v–t graph is shown in Fig. a.
sⴚt Graph: For the time interval 0 … t 6 9 s, the initial condition is s = 0 when
t = 0.
A+cB
ds = vdt
L0
s
ds =
s =
L0
t
4t3>2 dt
8 5>2
t
5
When t = 9 s,
s冷t = 9 s =
8 5>2
A 9 B = 388.8 m
5
For the time interval 9 s 6 t … 14 s, the initial condition is s = 388.8 m when
t = 9 s.
A+cB
ds = vdt
A 2t2 - 18t + 108 B dt
L388.8 m
L9 s
2
s = a t3 - 9t2 + 108t - 340.2b m
3
s
t
ds =
When t = 14 s,
s冷t = 14 s =
2
A 143 B - 9 A 142 B + 108(14) - 340.2 = 1237 m
3
The s–t graph is shown in Fig. b.
48
14
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*12–64. The jet bike is moving along a straight road
with the speed described by the v -s graph. Construct the
a-s graph.
v(m/s)
v ⫽ 5s1/2
75
v ⫽ ⫺0.2s ⫹ 120
15
a ⴚ s Graph: For 0 … s 6 225 m,
+ B
A:
a = v
s(m)
225
dv
5
= A 5s1>2 B a s - 1>2 b = 12.5 m>s2
ds
2
For 225 m 6 s … 525 m,
+ B
A:
a = v
dv
= (-0.2s + 120)(-0.2) = (0.04s - 24) m>s2
ds
At s = 225 m and 525 m,
a冷s = 225 m = 0.04(225) - 24 = -15 m>s2
a冷s = 525 m = 0.04(525) - 24 = -3 m>s2
The a–s graph is shown in Fig. a.
49
525
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•12–65. The acceleration of the speed boat starting from
rest is described by the graph. Construct the v-s graph.
a(ft/s2)
10
a ⫽ 0.04s ⫹ 2
2
N ⴚs Graph: For 0 … s 6 200 ft, the initial condition is v = 0 at s = 0.
+ B
A:
vdv = ads
L0
v
vdv =
L0
s
(0.04s + 2)ds
v
s
v2 2
= 0.02s2 + 2s冷0
2 0
v = 20.04s2 + 4s ft>s
At s = 200 ft,
v冷s = 200 ft = 20.04 A 2002 B + 4(200) = 48.99 ft>s = 49.0 ft >s
For 200 ft 6 s … 500 ft, the initial condition is v = 48.99 ft>s at s = 200 ft.
+ B
A:
vdv = ads
v
L48.99 ft>s
vdv =
s
10ds
L200 ft
v
s
v2 2
= 10s冷200 ft
2 48.99 ft>s
v = 220s - 1600 ft>s
At s = 500 ft,
v冷s = 500 ft = 220(500) - 1600 = 91.65 ft>s = 91.7 ft>s
The v–s graph is shown in Fig. a.
50
s(ft)
200
500
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12–66. The boat travels along a straight line with the
speed described by the graph. Construct the s-t and a-s
graphs. Also, determine the time required for the boat to
travel a distance s = 400 m if s = 0 when t = 0.
v(m/s)
80
sⴚt Graph: For 0 … s 6 100 m, the initial condition is s = 0 when t = 0 s.
A
+
:
B
v ⫽ 0.2s
ds
dt =
v
t
s
v2 ⫽ 4s
dt =
L0
L0 2s1>2
1>2
t = s
ds
20
s(m)
s = A t2 B m
100
When s = 100 m,
100 = t2
t = 10 s
For 100 6 s … 400 m, the initial condition is s = 100 m when t = 10 s.
+ B
A:
ds
v
dt =
t
s
L10 s
L100 m 0.2s
s
t - 10 = 5ln
100
t
s
- 2 = ln
5
100
s
et>5 - 2 =
100
s
et>5
=
100
e2
dt =
ds
s = A 13.53et>5 B m
When s = 400 m,
400 = 13.53et>5
t = 16.93 s = 16.9 s
Ans.
The s–t graph is shown in Fig. a.
aⴚs Graph: For 0 m … s 6 100 m,
a = v
dv
= A 2s1>2 B A s - 1>2 B = 2 m>s2
ds
For 100 m 6 s … 400 m,
a = v
dv
= (0.2s)(0.2) = 0.04s
ds
When s = 100 m and 400 m,
a冷s = 100 m = 0.04(100) = 4 m>s2
a冷s = 400 m = 0.04(400) = 16 m>s2
The a–s graph is shown in Fig. b.
51
400
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12–67. The s–t graph for a train has been determined
experimentally. From the data, construct the v-t and a–t
graphs for the motion.
s (m)
600
s ⫽ 24t ⫺ 360
360
s ⫽ 0.4t2
ds
.
Yⴚt Graph: The velocity in terms of time t can be obtained by applying y =
dt
For time interval 0 s … t … 30 s,
y =
When t = 30 s,
ds
= 0.8t
dt
y = 0.8(30) = 24.0 m>s
For time interval 30 s 6 t … 40 s,
y =
ds
= 24.0 m>s
dt
aⴚt Graph: The acceleration in terms of time t can be obtained by applying a =
dy
.
dt
dy
For time interval 0 s … t 6 30 s and 30 s 6 t … 40 s, a =
= 0.800 m>s2 and
dt
dy
a =
= 0, respectively.
dt
52
30
40
t (s)
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*12–68. The airplane lands at 250 ft>s on a straight
runway and has a deceleration described by the graph.
Determine the distance s¿ traveled before its speed is
decreased to 25 ft>s. Draw the s-t graph.
a(ft/s2)
1750
s
s(ft)
Nⴚs Graph: For 0 … s 6 1750 ft, the initial condition is v = 250 ft>s at s = 0 s.
⫺7.5
+ B
A:
⫺15
vdv = ads
v
s
vdv =
-15ds
L250 ft>s
L0
v
s
v2 2
= -15s冷0
2 250 ft>s
A 262 500 - 30s B ft>s
v =
At s = 1750 ft,
v冷s = 1750 ft = 262 500 - 30(1750) = 100 ft>s
For 1750 ft 6 s 6 s¿ , the initial condition is v = 100 ft>s at s = 1750 ft.
+ B
A:
vdv = ads
v
L100 ft>s
vdv =
s
L1750 ft
-7.5ds
v
s
v2 2
= (-7.5s)冷1750 ft
2 100 ft>s
v = 236 250 - 15s
When v = 25 ft>s
25 = 236 250 - 15s
s¿ = 2375 ft
Ans.
The v–s graph is shown in Fig. a.
53
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•12–69. The airplane travels along a straight runway with
an acceleration described by the graph. If it starts from rest
and requires a velocity of 90 m>s to take off, determine the
minimum length of runway required and the time t¿ for take
off. Construct the v-t and s-t graphs.
a(m/s2)
8
vⴚt graph: For the time interval 0 … t 6 10 s, the initial condition is v = 0 when
t = 0 s.
A
+
:
B
L0
dv =
L0
t
0.8tdt
v = A 0.4t2 B m>s
When t = 10 s,
v = 0.4 A 102 B = 40 m>s
For the time interval 10 s 6 t … t¿ , the initial condition is v = 40 m>s when t = 10 s.
+ B
A:
dv = adt
v
t
8dt
L10 s
L40 m>s
dv =
v
t
v冷40 m>s = 8t冷10 s
v = (8t - 40) m>s
Thus, when v = 90 m>s,
90 = 8t¿ - 40
t¿ = 16.25 s
Ans.
Also, the change in velocity is equal to the area under the a-t graph. Thus,
¢v =
L
t(s)
10
dv = adt
v
a ⫽ 0.8t
adt
1
(8)(10) + 8(t¿ - 10)
2
t¿ = 16.25 s
90 - 0 =
Ans.
The v–t graph is shown in Fig. a.
54
t
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sⴚt Graph: For the time interval 0 … t 6 10 s, the initial condition is s = 0 when
t = 0 s.
+ B
A:
ds = vdt
L0
s
ds =
L0
t
2
0.4t dt
s = A 0.1333t3 B m
When t = 10 s,
s冷t = 10 s = 0.1333 A 103 B = 133.33m
For the time interval 10 s 6 t … t¿ = 16.25 s, the initial condition is s = 133.33m
when t = 10 s.
+ B
A:
ds = vdt
s
L133.33 m
ds =
t
(8t - 40)dt
L10s
s冷133.33 m = A 4t2 - 40t B 2
s
t
10 s
s = A 4t - 40t + 133.33 B m
2
When t = t¿ = 16.25 s
s冷t = 16.25 s = 4(16.25)2 - 40(16.25) + 133.33 = 539.58 m = 540 m Ans.
The s–t graph is shown in Fig. b.
55
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12–70. The a–t graph of the bullet train is shown. If the
train starts from rest, determine the elapsed time t¿ before it
again comes to rest. What is the total distance traveled
during this time interval? Construct the v–t and s–t graphs.
a(m/s2)
Nⴚt Graph: For the time interval 0 … t 6 30 s, the initial condition is v = 0 when
t = 0 s.
+ B
A:
L0
dv =
L0
t
0.1tdt
v = A 0.05t2 B m>s
When t = 30 s,
v冷t = 30 s = 0.05 A 302 B = 45 m>s
For the time interval 30 s 6 t … t¿ , the initial condition is v = 45 m>s at t = 30 s.
+ B
A:
dv = adt
v
t
L45 m>s
dv =
v = ¢-
L30 s
¢-
1
t + 5 ≤ dt
15
1 2
t + 5t - 75 ≤ m>s
30
Thus, when v = 0,
0 = -
1 2
t¿ + 5t¿ - 75
30
Choosing the root t¿ 7 75 s,
t¿ = 133.09 s = 133 s
Ans.
Also, the change in velocity is equal to the area under the a–t graph. Thus,
¢v =
0 =
L
adt
1
1
1
(3)(75) + B ¢ - t¿ + 5 ≤ (t¿ - 75) R
2
2
15
0 = -
1 2
t¿ + 5t¿ - 75
30
56
1
a ⫽ ⫺( 15 )t ⫹ 5
t
30
dv = adt
v
a ⫽ 0.1t
3
75
t(s)
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This equation is the same as the one obtained previously.
The slope of the v–t graph is zero when t = 75 s, which is the instant a =
v冷t = 75 s = -
dv
= 0.Thus,
dt
1
A 752 B + 5(75) - 75 = 112.5 m>s
30
The v–t graph is shown in Fig. a.
sⴚt Graph: Using the result of v, the equation of the s–t graph can be obtained by
integrating the kinematic equation ds = vdt. For the time interval 0 … t 6 30 s, the
initial condition s = 0 at t = 0 s will be used as the integration limit. Thus,
+ B
A:
ds = vdt
L0
s
ds =
s = a
L0
t
0.05t2 dt
1 3
t bm
60
When t = 30 s,
1
A 303 B = 450 m
60
s冷t = 30 s =
For the time interval 30 s 6 t … t¿ = 133.09 s, the initial condition is s = 450 m
when t = 30 s.
+ B
A:
ds = vdt
s
L450 m
ds =
t
L30 s
a-
1 2
t + 5t - 75 bdt
30
5
1
s = a - t3 + t2 - 75t + 750b m
90
2
When t = 75 s and t¿ = 133.09 s,
s冷t = 75 s = -
1
5
A 753 B + A 752 B - 75(75) + 750 = 4500 m
90
2
s冷t = 133.09 s = -
1
5
A 133.093 B + A 133.092 B - 75(133.09) + 750 = 8857 m
90
2
The s–t graph is shown in Fig. b.
57
Ans.
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12–71. The position of a particle is r = 5(3t 3 - 2t)i
– (4t1/2 + t)j + (3t2 - 2)k6 m, where t is in seconds.
Determine the magnitude of the particle’s velocity and
acceleration when t = 2 s.
Velocity:
v =
dr
d
=
c A 3t3 - 2t B i - A 4t1>2 + t B j + A 3t2 - 2 B k d = c A 9t2 - 2 B i - A 2t - 1>2 + 1 B j + (6t)k d m>s
dt
dt
When t = 2 s,
v = B c9 A 22 B - 2 d i - c2 A 2 - 1>2 B + 1 dj + 6(2)k R m>s
= [34i - 2.414j + 12k] m>s
Thus, the magnitude of the particle’s velocity is
v = 2vx 2 + vy 2 + vz 2 = 2342 + (-2.414)2 + 122 = 36.1 m>s
Ans.
Acceleration:
a =
dv
d
=
c A 9t2 - 2 B i - A 2t - 1>2 + 1 B j + (6t)k d m>s = C (18t)i + t - 3>2j + 6k D m>s2
dt
dt
When t = 2 s,
a = C 18(2)i + 2 - 3>2j + 6k D m>s2 = [36i + 0.3536j + 6k] m>s2
Thus, the magnitude of the particle’s acceleration is
a = 2ax 2 + ay 2 + az 2 = 2362 + 0.35362 + 62 = 36.5 m>s2
Ans.
*12–72. The velocity of a particle is v = 53i + (6 - 2t)j6 m>s,
where t is in seconds. If r = 0 when t = 0, determine the
displacement of the particle during the time interval
t = 1 s to t = 3 s.
Position: The position r of the particle can be determined by integrating the
kinematic equation dr = vdt using the initial condition r = 0 at t = 0 as the
integration limit. Thus,
dr = vdt
L0
r
dr =
L0
t
C 3i + (6 - 2t)j D dt
r = c 3ti + A 6t - t2 B j dm
When t = 1 s and 3 s,
r冷t = 1 s = 3(1)i + C 6(1) - 12 D j = [3i + 5j] m>s
r冷t = 3 s = 3(3)i + C 6(3) - 32 D j = [9i + 9j] m>s
Thus, the displacement of the particle is
¢r = r冷t = 3 s - r冷t = 1 s
= (9i + 9j) - (3i + 5j)
= [6i + 4j] m
Ans.
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•12–73. A particle travels along the parabolic path
y = bx2. If its component of velocity along the y axis is
vy = ct2, determine the x and y components of the particle’s
acceleration. Here b and c are constants.
Velocity:
dy = vy dt
L0
y
dy =
y =
c 3
t
3
L0
t
ct2 dt
Substituting the result of y into y = bx2,
c 3
t = bx2
3
x =
c 3>2
t
A 3b
Thus, the x component of the particle’s velocity can be determined by taking the
time derivative of x.
d
c 3>2
3 c 1>2
#
vx = x =
t R =
t
B
dt A 3b
2A 3b
Acceleration:
3 c - 1>2
3 c 1
d 3 c 1>2
#
t ≤ =
t
=
ax = vx =
¢
dt 2A 3b
4A 3b
4A 3b 2t
Ans.
d 2
#
ay = vy =
A ct B = 2ct
dt
Ans.
59
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12–74. The velocity of a particle is given by
v = 516t2i + 4t3j + (5t + 2)k6 m>s, where t is in seconds. If
the particle is at the origin when t = 0, determine the
magnitude of the particle’s acceleration when t = 2 s. Also,
what is the x, y, z coordinate position of the particle at this
instant?
Acceleration: The acceleration expressed in Cartesian vector form can be obtained
by applying Eq. 12–9.
a =
dv
= {32ti + 12t2j + 5k} m>s2
dt
When t = 2 s, a = 32(2)i + 12 A 22 B j + 5k = {64i + 48j + 5k} m>s2. The magnitude
of the acceleration is
a = 2a2x + a2y + a2z = 2642 + 482 + 52 = 80.2 m>s2
Ans.
Position: The position expressed in Cartesian vector form can be obtained by
applying Eq. 12–7.
dr = v dt
L0
r
dr =
r = c
L0
t
A 16t2i + 4t3j + (5t + 2)k B dt
16 3
5
t i + t4j + a t2 + 2t bk d m
3
2
When t = 2 s,
r =
5
16 3
A 2 B i + A 24 B j + c A 22 B + 2(2) dk = {42.7i + 16.0j + 14.0k} m.
3
2
Thus, the coordinate of the particle is
(42.7, 16.0, 14.0) m
Ans.
60
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12–75. A particle travels along the circular path
x2 + y2 = r2. If the y component of the particle’s velocity is
vy = 2r cos 2t, determine the x and y components of its
acceleration at any instant.
Velocity:
dy = vy dt
L0
y
dy =
L0
t
2r cos 2tdt
y = r sin 2t
Substituting this result into x2 + y2 = r2, we obtain
x2 + r2 sin2 2t = r2
x2 = r2 A 1 - sin2 2t B
x = <r cos 2t
Thus,
d
#
(<r cos 2t) = <2r sin 2t
vx = x =
dt
Acceleration:
d
#
(<2r sin 2t) = <4r cos 2t
ax = vx =
dt
Ans.
d
#
ay = vy =
(2r cos 2t) = -4r sin 2t
dt
Ans.
61
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*12–76. The box slides down the slope described by the
equation y = (0.05x2) m, where x is in meters. If the box has
x components of velocity and acceleration of vx = –3 m>s
and ax = –1.5 m>s2 at x = 5 m, determine the y components
of the velocity and the acceleration of the box at this instant.
y
y ⫽ 0.05 x2
x
Velocity: The x and y components of the box’s velocity can be related by taking the
first time derivative of the path’s equation using the chain rule.
y = 0.05x2
#
#
y = 0.1xx
or
vy = 0.1xvx
At x = 5 m, vx = -3 m>s. Thus,
vy = 0.1(5)(-3) = -1.5 m>s = 1.5 m>s T
Ans.
Acceleration: The x and y components of the box’s acceleration can be obtained by
taking the second time derivative of the path’s equation using the chain rule.
# #
#
y = 0.1[xx + xx] = 0.1 A x2 + xx B
or
ay = 0.1 A vx 2 + xax B
At x = 5 m, vx = -3 m>s and ax = -1.5 m>s2. Thus,
ay = 0.1 C (-3)2 + 5(-1.5) D = 0.15 m>s2 c
Ans.
62
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•12–77. The position of a particle is defined by
r = 55 cos 2t i + 4 sin 2t j6 m, where t is in seconds and the
arguments for the sine and cosine are given in radians.
Determine the magnitudes of the velocity and acceleration
of the particle when t = 1 s. Also, prove that the path of the
particle is elliptical.
Velocity: The velocity expressed in Cartesian vector form can be obtained by
applying Eq. 12–7.
v =
dr
= {-10 sin 2ti + 8 cos 2tj} m>s
dt
When t = 1 s, v = -10 sin 2(1)i + 8 cos 2(1)j = {-9.093i - 3.329j} m>s. Thus, the
magnitude of the velocity is
y = 2y2x + y2y = 2( -9.093)2 + (-3.329)2 = 9.68 m>s
Ans.
Acceleration: The acceleration expressed in Cartesian vector from can be obtained
by applying Eq. 12–9.
a =
dv
= {-20 cos 2ti - 16 sin 2tj} m>s2
dt
When t = 1 s, a = -20 cos 2(1)i - 16 sin 2(1)j = {8.323i - 14.549j} m>s2. Thus, the
magnitude of the acceleration is
a = 2a2x + a2y = 28.3232 + (-14.549)2 = 16.8 m>s2
Ans.
Traveling Path: Here, x = 5 cos 2t and y = 4 sin 2t. Then,
x2
= cos2 2t
25
[1]
y2
= sin2 2t
16
[2]
Adding Eqs [1] and [2] yields
y2
x2
+
= cos2 2t + sin2 2t
25
16
However, cos2 2t + sin2 2t = 1. Thus,
y2
x2
+
= 1
25
16
(Equation of an Ellipse) (Q.E.D.)
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12–78. Pegs A and B are restricted to move in the elliptical
slots due to the motion of the slotted link. If the link moves
with a constant speed of 10 m/s, determine the magnitude of
the velocity and acceleration of peg A when x = 1 m.
y
A
C
D
x
v ⫽ 10 m/s
Velocity: The x and y components of the peg’s velocity can be related by taking the
first time derivative of the path’s equation.
x2
+ y2 = 1
4
1
#
#
(2xx) + 2yy = 0
4
1 #
#
xx + 2yy = 0
2
or
1
xv + 2yvy = 0
2 x
(1)
At x = 1 m,
(1)2
+ y2 = 1
4
y =
23
m
2
Here, vx = 10 m>s and x = 1. Substituting these values into Eq. (1),
1
23
(1)(10) + 2 ¢
≤ vy = 0
2
2
vy = -2.887 m>s = 2.887 m>s T
Thus, the magnitude of the peg’s velocity is
v = 2vx 2 + vy 2 = 2102 + 2.8872 = 10.4 m>s
Ans.
Acceleration: The x and y components of the peg’s acceleration can be related by
taking the second time derivative of the path’s equation.
1 # #
# #
(xx + xx) + 2(yy + yy) = 0
2
1 #2
#
A x + xx B + 2 A y 2 + yy B = 0
2
or
1
A v 2 + xax B + 2 A vy 2 + yay B = 0
2 x
Since nx is constant, ax = 0. When x = 1 m, y =
(2)
23
m, vx = 10 m>s, and
2
vy = -2.887 m>s. Substituting these values into Eq. (2),
23
1
a d = 0
A 102 + 0 B + 2 c(-2.887)2 +
2
2 y
ay = -38.49 m>s2 = 38.49 m>s2 T
Thus, the magnitude of the peg’s acceleration is
a = 2ax 2 + ay 2 = 202 + (-38.49)2 = 38.5 m>s2
Ans.
64
B
x2
⫹ y2 ⫽ 1
4
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12–79. A particle travels along the path y2 = 4x with a
constant speed of v = 4 m>s. Determine the x and y
components of the particle’s velocity and acceleration when
the particle is at x = 4 m.
Velocity: The x and y components of the particle’s velocity can be related by taking
the first time derivative of the path’s equation using the chain rule.
#
#
2yy = 4x
2 #
#
y = x
y
or
vy =
2
v
y x
(1)
1
v
2 x
(2)
At x = 4 m, y = 24(4) = 4 m. Thus Eq. (1) becomes
vy =
The magnitude of the particle’s velocity is
v = 2vx 2 + vy 2
(3)
Substituting v = 4 m>s and Eq. (2) into Eq. (3),
4 =
2
1
2
v
+
a
v
b
x
x
A
2
vx = 3.578 m>s = 3.58 m>s
Ans.
Substituting the result of nx into Eq. (2), we obtain
vy = 1.789 m>s = 1.79 m>s
Ans.
Acceleration: The x and y components of the particle’s acceleration can be related
by taking the second time derivative of the path’s equation using the chain rule.
# #
2(yy + yy) = 4x
#
y2 + yy = 2x
or
vy 2 + yay = 2ax
(4)
When x = 4 m, y = 4 m, and vy = 1.789 m>s. Thus Eq. (4) becomes
1.7892 + 4ay = 2ax
ay = 0.5ax - 0.8
(5)
Since the particle travels with a constant speed along the path, its acceleration along
the tangent of the path is equal to zero. Here, the angle that the tangent makes with the
dy
1
= tan - 1 ¢ 1>2 ≤ 2
= tan - 1 (0.5) = 26.57°.
horizontal at x = 4 m is u = tan - 1 ¢ ≤ 2
dx x = 4 m
x
x=4 m
Thus, from the diagram shown in Fig. a,
ax cos 26.57° + ay sin 26.57° = 0
(6)
Solving Eqs. (5) and (6) yields
ax = 0.32 m>s2
ay = -0.64 m>s2 = 0.64 m>s2 T
65
Ans.
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*12–80. The van travels over the hill described by
y = ( -1.5(10–3) x2 + 15) ft. If it has a constant speed of
75 ft>s, determine the x and y components of the van’s
velocity and acceleration when x = 50 ft.
y
15 ft
y ⫽ (⫺1.5 (10⫺3) x2 ⫹ 15) ft
x
Velocity: The x and y components of the van’s velocity can be related by taking the
first time derivative of the path’s equation using the chain rule.
y = -1.5 A 10 - 3 B x2 + 15
#
#
y = -3 A 10 - 3 B xx
or
vy = -3 A 10 - 3 B xvx
When x = 50 ft,
vy = -3 A 10 - 3 B (50)vx = -0.15vx
(1)
The magnitude of the van’s velocity is
v = 2vx 2 + vy 2
(2)
Substituting v = 75 ft>s and Eq. (1) into Eq. (2),
75 = 2vx 2 + (-0.15vx)2
vx = 74.2 ft>s ;
Ans.
Substituting the result of nx into Eq. (1), we obtain
vy = -0.15(-74.17) = 11.12 ft>s = 11.1 ft>s c
Ans.
Acceleration: The x and y components of the van’s acceleration can be related by
taking the second time derivative of the path’s equation using the chain rule.
$
# #
y = -3 A 10 - 3 B (xx + xx)
or
ay = -3 A 10 - 3 B A vx 2 + xax B
When x = 50 ft, vx = -74.17 ft>s. Thus,
ay = -3 A 10 - 3 B c(-74.17)2 + 50ax d
ay = -(16.504 + 0.15ax)
(3)
Since the van travels with a constant speed along the path, its acceleration along the tangent
of the path is equal to zero. Here, the angle that the tangent makes with the horizontal at
dy
x = 50 ft is u = tan - 1 ¢ ≤ 2
= tan - 1 c -3 A 10 - 3 B x d 2
= tan - 1( -0.15) = -8.531°.
dx x = 50 ft
x = 50 ft
Thus, from the diagram shown in Fig. a,
ax cos 8.531° - ay sin 8.531° = 0
(4)
Solving Eqs. (3) and (4) yields
ax = -2.42 ft>s = 2.42 ft>s2 ;
Ans.
ay = -16.1 ft>s = 16.1 ft>s2 T
Ans.
66
100 ft
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y
•12–81. A particle travels along the circular path from A
to B in 1 s. If it takes 3 s for it to go from A to C, determine
its average velocity when it goes from B to C.
30⬚
Position: The coordinates for points B and C are [30 sin 45°, 30 - 30 cos 45°] and
[30 sin 75°, 30 - 30 cos 75°]. Thus,
C
45⬚
30 m
rB = (30 sin 45° - 0)i + [(30 - 30 cos 45°) - 30]j
B
= {21.21i - 21.21j} m
A
rC = (30 sin 75° - 0)i + [(30 - 30 cos 75°) - 30]j
= {28.98i - 7.765j} m
Average Velocity: The displacement from point B to C is ¢rBC = rC - rB
= (28.98i - 7.765j) - (21.21i - 21.21j) = {7.765i + 13.45j} m.
(vBC)avg =
7.765i + 13.45j
¢rBC
=
= {3.88i + 6.72j} m>s
¢t
3 - 1
Ans.
12–82. A car travels east 2 km for 5 minutes, then north
3 km for 8 minutes, and then west 4 km for 10 minutes.
Determine the total distance traveled and the magnitude
of displacement of the car. Also, what is the magnitude of
the average velocity and the average speed?
Total Distance Traveled and Displacement: The total distance traveled is
s = 2 + 3 + 4 = 9 km
Ans.
and the magnitude of the displacement is
¢r = 2(2 + 4)2 + 32 = 6.708 km = 6.71 km
Ans.
Average Velocity and Speed: The total time is ¢t = 5 + 8 + 10 = 23 min = 1380 s.
The magnitude of average velocity is
yavg =
6.708 A 103 B
¢r
=
= 4.86 m>s
¢t
1380
and the average speed is
A ysp B avg
Ans.
9 A 103 B
s
=
=
= 6.52 m>s
¢t
1380
Ans.
67
x
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12–83. The roller coaster car travels down the helical path at
constant speed such that the parametric equations that define
its position are x = c sin kt, y = c cos kt, z = h - bt, where
c, h, and b are constants. Determine the magnitudes of its
velocity and acceleration.
z
x = c sin kt
#
x = ck cos kt
$
x = -ck2 sin kt
y = c cos kt
#
y = -ck sin kt
$
y = -ck2 cos kt
z = h - bt
#
z = -b
$
z = 0
y
v = 2(ck cos kt)2 + (-ck sin kt)2 + (-b)2 = 2c2k2 + b2
x
Ans.
a = 2(-ck2 sin kt)2 + (-ck2 cos kt)2 + 0 = ck2
Ans.
*12–84. The path of a particle is defined by y2 = 4kx, and
the component of velocity along the y axis is vy = ct, where
both k and c are constants. Determine the x and y
components of acceleration when y = y0.
y2 = 4kx
2yvy = 4kvx
2v2y + 2yay = 4kax
vy = ct
ay = c
Ans.
2(ct)2 + 2yc = 4kax
ax =
c
A y + ct2 B
2k
Ans.
68
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•12–85. A particle moves along the curve y = x - (x2>400),
where x and y are in ft. If the velocity component in the x
direction is vx = 2 ft>s and remains constant, determine the
magnitudes of the velocity and acceleration when x = 20 ft.
Velocity: Taking the first derivative of the path y = x -
x2
, we have
400
1
#
#
#
y = x (2xx)
400
x #
#
#
y = x x
200
[1]
#
#
However, x = yx and y = yy. Thus, Eq. [1] becomes
yy = yx -
x
y
200 x
[2]
Here, yx = 2 ft>s at x = 20 ft. Then, From Eq. [2]
yy = 2 Also,
20
(2) = 1.80 ft>s
200
y = 2y2x + y2y = 222 + 1.802 = 2.69 ft>s
Acceleration: Taking the second derivative of the path y = x -
Ans.
x2
, we have
400
1 #2
$
$
$
y = x A x + xx B
200
[3]
$
$
However, x = ax and y = ay. Thus, Eq. [3] becomes
ay = ax -
1
A y2 + xax B
200 x
[4]
Since yx = 2 ft>s is constant, hence ax = 0 at x = 20 ft. Then, From Eq. [4]
ay = 0 Also,
1
C 22 + 20(0) D = -0.020 ft>s2
200
a = 2a2x + a2y = 202 + ( -0.020)2 = 0.0200 ft>s2
Ans.
69
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12–86. The motorcycle travels with constant speed v0
along the path that, for a short distance, takes the form of a
sine curve. Determine the x and y components of its
velocity at any instant on the curve.
y
v0
π x)
y ⫽ c sin ( ––
L
x
p
y = c sin a xb
L
y =
c
c
L
L
p
p
#
cacos xbx
L
L
vy =
p
p
c vx acos xb
L
L
v20 = v2y + v2x
v20 = v2x B 1 + a
p 2
p
cb cos2 a xb R
L
L
vx = v0 B 1 + a
-2
p 2
p
cb cos2 a xb R
L
L
1
Ans.
1
vy =
-2
v0 pc
p
p
p 2
a cos xb B 1 + a cb cos2 a xb R
L
L
L
L
Ans.
12–87. The skateboard rider leaves the ramp at A with an
initial velocity vA at a 30° angle. If he strikes the ground at
B, determine vA and the time of flight.
vA
A
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
x-Motion: Here, (vA)x = vA cos 30°, xA = 0 and xB = 5 m. Thus,
+ B
A:
xB = xA + (vA)xt
5 = 0 + vA cos 30° t
t =
5
vA cos 30°
(1)
y-Motion: Here, (vA)y = vA sin 30°, ay = -g = -9.81 m>s2, and yB = -1 m. Thus,
A+cB
yB = yA + (vA)y t +
1
a t2
2 y
-1 = 0 + vA sin 30° t +
1
(-9.81)t2
2
4.905t2 - vA sin 30° t - 1 = 0
(2)
Solving Eqs. (1) and (2) yields
vA = 6.49 m>s
30⬚
1m
t = 0.890 s
Ans.
70
B
5m
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*12–88. The pitcher throws the baseball horizontally with
a speed of 140 ft>s from a height of 5 ft. If the batter is 60 ft
away, determine the time for the ball to arrive at the batter
and the height h at which it passes the batter.
A
+
;
B
60 ft
s = vt;
60 = 140t
t = 0.4286 = 0.429 s
A+cB
5 ft
h
s = s0 + v0 t +
Ans.
1
a t2
2 c
h = 5 + 0 +
1
(-32.2)(0.4286)2 = 2.04 ft
2
Ans.
•12–89. The ball is thrown off the top of the building. If it
strikes the ground at B in 3 s, determine the initial velocity vA
and the inclination angle uA at which it was thrown. Also, find
the magnitude of the ball’s velocity when it strikes the ground.
vA
uA
A
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
75 ft
x-Motion: Here, (vA)x = vA cos u, xA = 0, and xB = 60 ft, and t = 3 s. Thus,
+ B
A:
xB = xA + (vA)xt
60 = 0 + vA cos u(3)
vA cos u = 20
(1)
2
y-Motion: Here, (vA)y = vA sin u, ay = -g = -32.2 ft>s , yA = 0, and yB = -75 ft,
and t = 3 s. Thus,
A+cB
1 2
a t
2 y
1
-75 = 0 + vA sin u(3) + (-32.2) A 32 B
2
yB = yA + (vA)y t +
vA sin u = 23.3
(2)
Solving Eqs. (1) and (2) yields
u = 49.36° = 49.4°
vA = 30.71 ft>s = 30.7 ft>s
Ans.
Using the result of u and nA, we obtain
(vA)x = 30.71 cos 49.36° = 20 ft>s
(vA)y = 30.71 sin 49.36° = 23.3 ft>s
Thus,
A+cB
(vB)y = (vA)y + ayt
(vB)y = 23.3 + (-32.2)(3) = -73.3 ft>s = 73.3 ft>s T
Thus, the magnitude of the ball’s velocity when it strikes the ground is
vB = 2202 + 73.32 = 76.0 ft>s
Ans.
71
B
60 ft
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12–90. A projectile is fired with a speed of v = 60 m>s at
an angle of 60°. A second projectile is then fired with the
same speed 0.5 s later. Determine the angle u of the second
projectile so that the two projectiles collide. At what
position (x, y) will this happen?
y
v ⫽ 60 m/s
60⬚
u
v ⫽ 60 m/s
x
x
x-Motion: For the motion of the first projectile, vx = 60 cos 60° = 30 m>s, x0 = 0,
and t = t1. Thus,
+ B
A:
x = x0 + vxt
x = 0 + 30t1
(1)
For the motion of the second projectile, vx = 60 cos u, x0 = 0, and t = t1 - 0.5.
Thus,
+ B
A:
x = x0 + vxt
x = 0 + 60 cos u(t1 - 0.5)
(2)
y-Motion: For the motion of the first projectile, vy = 60 sin 60° = 51.96 m>s, y0 = 0,
and ay = -g = -9.81 m>s2. Thus,
A+cB
1 2
at
2 y
1
y = 0 + 51.96t1 + (-9.81)t1 2
2
y = y0 + vyt +
y = 51.96t1 - 4.905t1 2
For the motion of the
ay = -g = -9.81 m>s2. Thus,
A+cB
y = y0 + vyt +
second
(3)
projectile,
vy = 60 sin u,
y0 = 0,
and
1 2
at
2 y
y = 0 + 60 sin u(t1 - 0.5) +
1
(-9.81)(t1 - 0.5)2
2
y = (60 sin u)t1 - 30 sin u - 4.905 t1 2 + 4.905t1 - 1.22625
(4)
Equating Eqs. (1) and (2),
30t1 = 60 cos u(t1 - 0.5)
t1 =
cos u
2 cos u - 1
(5)
Equating Eqs. (3) and (4),
51.96t1 - 4.905t1 2 = (60 sin u)t1 - 30 sin u - 4.905t1 2 + 4.905t1 - 1.22625
(60 sin u - 47.06)t1 = 30 sin u + 1.22625
t1 =
30 sin u + 1.22625
60 sin u - 47.06
(6)
72
y
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Equating Eqs. (5) and (6) yields
30 sin u + 1.22625
cos u
=
2 cos u - 1
60 sin u - 47.06
49.51 cos u - 30 sin u = 1.22625
Solving by trial and error,
u = 57.57° = 57.6°
Ans.
Substituting this result into Eq. (5) (or Eq. (6)),
t1 =
cos 57.57°
= 7.3998 s
2 cos 57.57° - 1
Substituting this result into Eqs. (1) and (3),
x = 30(7.3998) = 222 m
Ans.
y = 51.96(7.3998) - 4.905 A 7.39982 B = 116 m
Ans.
73
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12–91. The fireman holds the hose at an angle u = 30°
with horizontal, and the water is discharged from the hose
at A with a speed of vA = 40 ft>s. If the water stream strikes
the building at B, determine his two possible distances s
from the building.
vA ⫽ 40 ft/s
A
B
u
8 ft
4 ft
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
x-Motion: Here, (vA)x = 40 cos 30° ft>s = 34.64 ft>s, xA = 0, and xB = s. Thus,
+ B
A:
xB = xA + (vA)xt
s = 0 + 34.64t
s = 34.64t
(1)
y-Motion: Here, (vA)y = 40 sin 30° ft>s = 20 ft>s, ay = -g = -32.2 ft>s2, yA = 0,
and yB = 8 - 4 = 4 ft. Thus,
A+cB
yB = yA + (vA)y t +
4 = 0 + 20t +
1
a t2
2 y
1
(-32.2)t2
2
16.1t2 - 20t + 4 = 0
t = 0.2505 s and 0.9917 s
Substituting these results into Eq. (1), the two possible distances are
s = 34.64(0.2505) = 8.68 ft
Ans.
s = 34.64(0.9917) = 34.4 ft
Ans.
74
s
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*12–92. Water is discharged from the hose with a speed of
40 ft>s. Determine the two possible angles u the fireman can
hold the hose so that the water strikes the building at B.
Take s = 20 ft.
vA ⫽ 40 ft/s
A
B
u
8 ft
4 ft
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
x-Motion: Here, (vA)x = 40 cos u, xA = 0, and xB = 20 ft>s. Thus,
+ B
A:
xB = xA + (vA)xt
20 = 0 + 40 cos ut
t =
1
2 cos u
(1)
y-Motion: Here, (vA)y = 40 sin u, ay = -g = -32.2 ft>s2, yA = 0, and yB = 8 - 4
= 4 ft. Thus,
A+cB
1
a t2
2 y
1
4 = 0 + 40 sin ut + (-32.2)t2
2
yB = yA + (vA)yt +
16.1t2 - 40 sin ut + 4 = 0
(2)
Substituting Eq. (1) into Eq. (2) yields
16.1 ¢
2
1
1
≤ - 40 sin u ¢
≤ + 4 = 0
2 cos u
2 cos u
20 sin u cos u - 4 cos2 u = 4.025
10 sin u cos u - 2 cos2 u = 2.0125
5 sin 2u - A 2 cos2 u - 1 B - 1 = 2.0125
5 sin 2u - cos 2u = 3.0125
Solving by trial and error,
u = 23.8° and 77.5°
Ans.
75
s
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•12–93. The pitching machine is adjusted so that the
baseball is launched with a speed of vA = 30 m>s. If the ball
strikes the ground at B, determine the two possible angles uA
at which it was launched.
vA ⫽ 30 m/s
A
uA
1.2 m
B
30 m
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
x-Motion: Here, (vA)x = 30 cos uA, xA = 0 and xB = 30 m. Thus,
+ B
A:
xB = xA + (vA)xt
30 = 0 + 30 cos uAt
t =
1
cos uA
(1)
y-Motion: Here, (vA)y = 30 sin uA, ay = -g = -9.81 m>s2, and yB = -1.2 m. Thus,
A+cB
1
a t2
2 y
1
-1.2 = 0 + 30 sin uAt + (-9.81)t2
2
yB = yA + (vA)yt +
4.905t2 - 30 sin uA t - 1.2 = 0
(2)
Substituting Eq. (1) into Eq. (2) yields
4.905 ¢
2
1
1
≤ - 30 sin uA ¢
≤ - 1.2 = 0
cos uA
cos uA
1.2 cos2 uA + 30 sin uA cos uA - 4.905 = 0
Solving by trial and error,
uA = 7.19° and 80.5°
Ans.
76
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12–94. It is observed that the time for the ball to strike the
ground at B is 2.5 s. Determine the speed vA and angle uA at
which the ball was thrown.
vA
uA
A
1.2 m
B
50 m
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
x-Motion: Here, (vA)x = vA cos uA, xA = 0, xB = 50 m, and t = 2.5 s. Thus,
+ B
A:
xB = xA + (vA)xt
50 = 0 + vA cos uA(2.5)
vA cos uA = 20
(1)
y-Motion: Here, (vA)y = vA sin uA,
= -9.81 m>s2. Thus,
A+cB
yB = yA + (vA)y t +
yA = 0 ,
yB = -1.2 m,
and
ay = -g
1
a t2
2 y
-1.2 = 0 + vA sin uA (2.5) +
1
(-9.81) A 2.52 B
2
vA sin uA = 11.7825
(2)
Solving Eqs. (1) and (2) yields
vA = 23.2 m>s
uA = 30.5°
Ans.
12–95. If the motorcycle leaves the ramp traveling at
110 ft>s, determine the height h ramp B must have so that
the motorcycle lands safely.
110 ft/s
30⬚
30 ft
A
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with the take off point of the motorcycle at ramp A.
x-Motion: Here, xA = 0, xB = 350 ft, and (vA)x = 110 cos 30° = 95.26 ft>s. Thus,
+ B
A:
xB = xA + (vA)xt
350 = 0 + 95.26t
t = 3.674 s
y-Motion: Here, yA = 0, yB = h - 30, (vA)y = 110 sin 30° = 55 ft>s, and ay = -g
= -32.2 ft>s2. Thus, using the result of t, we have
A+cB
yB = yA + (vA)yt +
1
a t2
2 y
h - 30 = 0 + 55(3.674) +
1
(-32.2) A 3.6742 B
2
h = 14.7 ft
Ans.
77
h
350 ft
B
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*12–96. The baseball player A hits the baseball with
vA = 40 ft>s and uA = 60°. When the ball is directly above
of player B he begins to run under it. Determine the
constant speed vB and the distance d at which B must run in
order to make the catch at the same elevation at which the
ball was hit.
vA ⫽ 40 ft/s
A
uA
15 ft
Vertical Motion: The vertical component of initial velocity for the football is
(y0)y = 40 sin 60° = 34.64 ft>s. The initial and final vertical positions are (s0)y = 0
and sy = 0, respectively.
(+ c )
sy = (s0)y + (y0)y t +
0 = 0 + 34.64t +
1
(a ) t2
2 cy
1
(-32.2)t2
2
t = 2.152 s
Horizontal Motion: The horizontal component of velocity for the baseball is
(y0)x = 40 cos 60° = 20.0 ft>s. The initial and final horizontal positions are
(s0)x = 0 and sx = R, respectively.
+ B
A:
sx = (s0)x + (y0)x t
R = 0 + 20.0(2.152) = 43.03 ft
The distance for which player B must travel in order to catch the baseball is
d = R - 15 = 43.03 - 15 = 28.0 ft
Ans.
Player B is required to run at a same speed as the horizontal component of velocity
of the baseball in order to catch it.
yB = 40 cos 60° = 20.0 ft>s
Ans.
78
B
C
vB
d
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•12–97. A boy throws a ball at O in the air with a speed v0
at an angle u1. If he then throws another ball with the same
speed v0 at an angle u2 6 u1, determine the time between
the throws so that the balls collide in mid air at B.
B
O
Vertical Motion: For the first ball, the vertical component of initial velocity is
(y0)y = y0 sin u1 and the initial and final vertical positions are (s0)y = 0 and sy = y,
respectively.
(+ c )
sy = (s0)y + (y0)y t +
1
(a ) t2
2 cy
y = 0 + y0 sin u1t1 +
1
(-g)t21
2
[1]
For the second ball, the vertical component of initial velocity is (y0)y = y0 sin u2 and
the initial and final vertical positions are (s0)y = 0 and sy = y, respectively.
(+ c )
sy = (s0)y + (y0)y t +
1
(a ) t2
2 cy
y = 0 + y0 sin u2t2 +
1
(-g)t22
2
[2]
Horizontal Motion: For the first ball, the horizontal component of initial velocity is
(y0)x = y0 cos u1 and the initial and final horizontal positions are (s0)x = 0 and
sx = x, respectively.
+ B
A:
sx = (s0)x + (y0)x t
x = 0 + y0 cos u1 t1
[3]
For the second ball, the horizontal component of initial velocity is (y0)x = y0 cos u2
and the initial and final horizontal positions are (s0)x = 0 and sx = x, respectively.
+ B
A:
sx = (s0)x + (y0)x t
x = 0 + y0 cos u2 t2
[4]
Equating Eqs. [3] and [4], we have
t2 =
cos u1
t
cos u2 1
[5]
Equating Eqs. [1] and [2], we have
y0 t1 sin u1 - y0 t2 sin u2 =
1
g A t21 - t22 B
2
[6]
Solving Eq. [5] into [6] yields
t1 =
t2 =
2y0 cos u2 sin(u1 - u2)
g(cos2 u2 - cos2 u1)
2y0 cos u1 sin(u1 - u2)
g(cos2 u2 - cos2u1)
Thus, the time between the throws is
¢t = t1 - t2 =
=
2y0 sin(u1 - u2)(cos u2 - cos u1)
g(cos2 u2 - cos2 u1)
2y0 sin (u1 - u2)
g(cos u2 + cos u1)
Ans.
79
u1
y
u2
x
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12–98. The golf ball is hit at A with a speed of
vA = 40 m>s and directed at an angle of 30° with the
horizontal as shown. Determine the distance d where the
ball strikes the slope at B.
B
vA ⫽ 40 m/s
30⬚
A
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
5
x-Motion: Here, (vA)x = 40 cos 30° = 34.64 m>s, xA = 0, and xB = d ¢
≤
2
25
+ 1
= 0.9806d. Thus,
+ B
A:
xB = xA + (vA)xt
0.9806d = 0 + 34.64t
t = 0.02831d
(1)
y-Motion: Here, (vA)y = 40 sin 30° = 20 m>s, yA = 0, yB = d ¢
= 0.1961d, and ay = -g = -9.81 m>s2.
1
≤
A 252 + 1
Thus,
A+cB
1
a t2
2 y
1
0.1961d = 0 + 20t + (-9.81)t2
2
yB = yA + (vA)yt +
4.905t2 - 20t + 0.1961d = 0
(2)
Substituting Eq. (1) into Eq. (2) yields
4.905(0.02831d)2 - 20(0.02831d) + 0.1961d = 0
3.9303 A 10 - 3 B d2 - 0.37002d = 0
d C 3.9303 A 10 - 3 B d - 0.37002 D = 0
Since d Z 0, then
3.9303 A 10 - 3 B d = 0.37002 = 0
d = 94.1 m
Ans.
80
5
d
1
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12–99. If the football is kicked at the 45° angle, determine
its minimum initial speed vA so that it passes over the goal
post at C. At what distance s from the goal post will the
football strike the ground at B?
vA
45⬚
A
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
x-Motion: For the motion from A to C, xA = 0, and xC = 160 ft, (vA)x = vA cos 45°,
and t = tAC. Thus,
+ B
A:
xC = xA + (vA)xt
160 = 0 + vA cos 45° tAC
tAC =
160
vA cos 45°
(1)
For the motion from A to B, xA = 0, and xB = 160 + s, (vA)x = vA cos 45°, and
t = tAB. Thus,
+ B
A:
xB = xA + (vA)xt
160 + s = 0 + vA cos 45° tAB
s = vA cos 45°(tAB) - 160
(2)
y-Motion: For the motion from A to C, yA = 0, and yC = 20 ft, (vA)y = vA sin 45°,
and ay = -g = -32.2 ft>s2. Thus,
A+cB
yC = yA + (vA)yt +
1
a t2
2 y
20 = 0 + vA sin 45° tAC +
1
(-32.2)tAC 2
2
16.1tAC 2 - vA sin 45° tAC + 20 = 0
(3)
For the motion from A to B, yA = yB = 0. Thus,
A+cB
yB = yA + (vA)yt +
1
a t2
2 y
0 = 0 + vA sin 45°(tAB) +
1
(-32.2)tAB 2
2
tAB (16.1tAB - vA sin 45°) = 0
Since tAB Z 0, then
16.1tAB - vA sin 45° = 0
(4)
Substituting Eq. (1) into Eq. (3) yields
16.1 ¢
C
20 ft
2
160
160
≤ - vA sin 45° ¢
≤ + 20 = 0
vA cos 45°
vA cos 45°
vA = 76.73 ft>s = 76.7 ft>s
Ans.
Substituting this result into Eq. (4),
16.1tAB - 76.73 sin 45° = 0
tAB = 3.370 s
Substituting the result of tAB and yA into Eq. (2),
s = 76.73 cos 45°(3.370) - 160
= 22.9 ft
Ans.
81
160 ft
s
B
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*12–100. The velocity of the water jet discharging from the
orifice can be obtained from v = 22 gh, where h = 2 m is
the depth of the orifice from the free water surface. Determine
the time for a particle of water leaving the orifice to reach
point B and the horizontal distance x where it hits the surface.
2m
A
vA
1.5 m
B
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A. The speed of the water that the jet discharges from A is
x
vA = 22(9.81)(2) = 6.264 m>s
x-Motion: Here, (vA)x = vA = 6.264 m>s, xA = 0, xB = x, and t = tA. Thus,
+ B
A:
xB = xA + (vA)xt
x = 0 + 6.264tA
(1)
y-Motion: Here, (vA)y = 0, ay = -g = -9.81 m>s2, yA = 0 m, yB = -1.5 m, and
t = tA. Thus,
A+cB
yB = yA + (vA)yt +
-1.5 = 0 + 0 +
1
a t2
2 y
1
(-9.81)tA 2
2
tA = 0.553 s
Thus,
x = 0 + 6.264(0.553) = 3.46 m
Ans.
•12–101. A projectile is fired from the platform at B. The
shooter fires his gun from point A at an angle of 30°.
Determine the muzzle speed of the bullet if it hits the
projectile at C.
B
C
vA
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
xC = xA + (vA)xt
20 = 0 + vA cos 30° t
(1)
y-Motion: Here, yA = 1.8, (vA)y = vA sin 30°, and ay = -g = -9.81 m>s2. Thus,
A+cB
yC = yA + (vA)yt +
1
a t2
2 y
10 = 1.8 + vA sin 30°(t) +
1
(-9.81)(t)2
2
Thus,
10 - 1.8 = ¢
20 sin 30°
≤ (t) - 4.905(t)2
cos 30°(t)
t = 0.8261 s
So that
vA =
10 m
30⬚
1.8 m
20 m
x-Motion: Here, xA = 0 and xC = 20 m. Thus,
+ B
A:
A
20
= 28.0 m>s
cos 30°(0.8261)
Ans.
82
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12–102. A golf ball is struck with a velocity of 80 ft>s as
shown. Determine the distance d to where it will land.
Horizontal Motion: The horizontal component of velocity is (y0)x = 80 cos 55°
= 45.89 ft>s.The initial and final horizontal positions are (s0)x = 0 and sx = d cos 10°,
respectively.
A
+
:
B
vA ⫽ 80 ft/s
B
A
45⬚
10⬚
d
sx = (s0)x + (y0)x t
d cos 10° = 0 + 45.89t
[1]
Vertical Motion: The vertical component of initial velocity is (y0)y = 80 sin 55°
= 65.53 ft>s. The initial and final vertical positions are (s0)y = 0 and sy = d sin 10°,
respectively.
(+ c )
1
(a ) t2
2 cy
1
d sin 10° = 0 + 65.53t + (-32.2)t2
2
sy = (s0)y + (y0)y t +
[2]
Solving Eqs. [1] and [2] yields
d = 166 ft
Ans.
t = 3.568 s
12–103. The football is to be kicked over the goalpost,
which is 15 ft high. If its initial speed is vA = 80 ft>s,
determine if it makes it over the goalpost, and if so, by how
much, h.
h
vA ⫽ 80 ft/s
B
15 ft
60⬚
Horizontal Motion: The horizontal component of velocity is (y0)x = 80 cos 60°
= 40.0 ft>s. The initial and final horizontal positions are (s0)x = 0 and sx = 25 ft,
respectively.
+ B
A:
sx = (s0)x + (y0)x t
25 = 0 + 40.0t
t = 0.625 s
Vertical Motion: The vertical component of initial velocity is (y0)y = 80 sin 60°
= 69.28 ft>s. The initial and final vertical positions are (s0)y = 0 and sy = H,
respectively.
A+cB
sy = (s0)y + (y0)y t +
H = 0 + 69.28(0.625) +
1
(a ) t2
2 cy
1
(-32.2) A 0.6252 B
2
H = 37.01 ft
Since H 7 15 ft , the football is kicked over the goalpost.
Ans.
h = H - 15 = 37.01 - 15 = 22.0 ft
Ans.
83
25 ft
45⬚
30 ft
x
y
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*12–104. The football is kicked over the goalpost with an
initial velocity of vA = 80 ft>s as shown. Determine the
point B (x, y) where it strikes the bleachers.
h
vA ⫽ 80 ft/s
Horizontal Motion: The horizontal component of velocity is (y0)x = 80 cos 60°
= 40.0 ft>s. The initial and final horzontal positions are (s0)x = 0 and sx = (55 + x),
respectively.
+ B
A:
B
15 ft
60⬚
45⬚
25 ft
x
30 ft
sx = (s0)x + (y0)x t
55 + x = 0 + 40.0t
[1]
Vertical Motion: The vertical component of initial velocity is (y0)y = 80 sin 60°
= 69.28 ft>s. The initial and final vertical positions are (s0)y = 0 and
sy = y = x tan 45° = x, respectively.
A+cB
sy = (s0)y + (y0)y t +
x = 0 + 69.28t +
1
(a ) t2
2 cy
1
(-32.2)t2
2
[2]
Solving Eqs. [1] and [2] yields
t = 2.969 s
y = x = 63.8 ft
Ans.
•12–105. The boy at A attempts to throw a ball over the
roof of a barn with an initial speed of vA = 15 m>s.
Determine the angle uA at which the ball must be thrown so
that it reaches its maximum height at C. Also, find the
distance d where the boy should stand to make the throw.
C
8m
vA
A
Vertical Motion: The vertical component, of initial and final velocity are
(y0)y = (15 sin uA) m>s and yy = 0, respectively. The initial vertical position is
(s0)y = 1 m.
A+cB
A+cB
yy = (y0) + ac t
0 = 15 sin uA + (-9.81)t
[1]
1
(a ) t2
2 cy
1
8 = 1 + 15 sin uAt + (-9.81)t2
2
sy = (s0)y + (y0)y t +
[2]
Solving Eqs. [1] and [2] yields
uA = 51.38° = 51.4°
Ans.
t = 1.195 s
Horizontal Motion: The horizontal component of velocity is (y0)x = yA cos uA
= 15 cos 51.38° = 9.363 m>s. The initial and final horizontal positions are (s0)x = 0
and sx = (d + 4) m, respectively.
+ B
A:
sx = (s0)x + (y0)x t
d + 4 = 0 + 9.363(1.195)
d = 7.18 m
Ans.
84
uA
1m
d
4m
y
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12–106. The boy at A attempts to throw a ball over the
roof of a barn such that it is launched at an angle uA = 40°.
Determine the minimum speed vA at which he must throw
the ball so that it reaches its maximum height at C. Also,
find the distance d where the boy must stand so that he can
make the throw.
C
8m
vA
uA
A
1m
Vertical Motion: The vertical components of initial and final velocity are
(y0)y = (yA sin 40°) m>s and yy = 0, respectively. The initial vertical position is
(s0)y = 1 m.
A+cB
4m
yy = (y0) + ac t
0 = yA sin 40° + (-9.81) t
A+cB
d
[1]
1
(a ) t2
2 cy
1
8 = 1 + yA sin 40°t + (-9.81) t2
2
sy = (s0)y + (y0)y t +
[2]
Solving Eqs. [1] and [2] yields
yA = 18.23 m>s = 18.2 m>s
Ans.
t = 1.195 s
Horizontal Motion: The horizontal component of velocity is (y0)x = yA cos uA
= 18.23 cos 40° = 13.97 m>s. The initial and final horizontal positions are (s0)x = 0
and sx = (d + 4) m, respectively.
+ B
A:
sx = (s0)x + (y0)x t
d + 4 = 0 + 13.97(1.195)
d = 12.7 m
Ans.
12–107. The fireman wishes to direct the flow of water
from his hose to the fire at B. Determine two possible
angles u1 and u2 at which this can be done. Water flows from
the hose at vA = 80 ft>s.
+ B
A:
A
u
vA
20 ft
s = s0 + v0 t
B
35 = 0 + (80) cos u
A+cB
s = s0 + v0 t +
1 2
at
2 c
-20 = 0 - 80 sin u t +
35 ft
1
( -32.2)t2
2
Thus,
20 = 80 sin u
0.4375
0.1914
t + 16.1 ¢
≤
cos u
cos2 u
20 cos2 u = 17.5 sin 2u + 3.0816
Solving,
u1 = 25.0° (below the horizontal)
Ans.
u2 = 85.2° (above the horizontal)
Ans.
85
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*12–108. Small packages traveling on the conveyor belt
fall off into a l-m-long loading car. If the conveyor is running
at a constant speed of vC = 2 m>s, determine the smallest
and largest distance R at which the end A of the car may be
placed from the conveyor so that the packages enter the car.
vc ⫽ 2 m/s
30⬚
3m
A
Vertical Motion: The vertical component of initial velocity is (y0)y = 2 sin 30°
= 1.00 m>s. The initial and final vertical positions are (s0)y = 0 and sy = 3 m,
respectively.
A+TB
sy = (s0)y + (y0)y t +
3 = 0 + 1.00(t) +
1
(a ) t2
2 cy
1
(9.81) A t2 B
2
Choose the positive root t = 0.6867 s
Horizontal Motion: The horizontal component of velocity is (y0)x = 2 cos 30°
= 1.732 m>s and the initial horizontal position is (s0)x = 0. If sx = R, then
+ B
A:
sx = (s0)x + (y0)x t
R = 0 + 1.732(0.6867) = 1.19 m
Ans.
If sx = R + 1, then
+ B
A:
sx = (s0)x + (y0)x t
R + 1 = 0 + 1.732(0.6867)
R = 0.189 m
Ans.
86
R
B
1m
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•12–109. Determine the horizontal velocity vA of a tennis
ball at A so that it just clears the net at B. Also, find the
distance s where the ball strikes the ground.
vA
B
s
Vertical Motion: The vertical component of initial velocity is (y0)y = 0. For the ball
to travel from A to B, the initial and final vertical positions are (s0)y = 7.5 ft and
sy = 3 ft, respectively.
A+cB
sy = (s0)y + (y0)y t +
3 = 7.5 + 0 +
1
(a ) t2
2 cy
1
( -32.2)t 21
2
t1 = 0.5287 s
For the ball to travel from A to C, the initial and final vertical positions are
(s0)y = 7.5 ft and sy = 0, respectively.
A+cB
sy = (s0)y + (y0)y t +
0 = 7.5 + 0 +
1
(a ) t2
2 cy
1
( -32.2)t22
2
t2 = 0.6825 s
Horizontal Motion: The horizontal component of velocity is (y0)x = yA. For the ball
to travel from A to B, the initial and final horizontal positions are (s0)x = 0 and
sx = 21 ft, respectively. The time is t = t1 = 0.5287 s.
+ B
A;
sx = (s0)x + (y0)x t
21 = 0 + yA (0.5287)
yA = 39.72 ft>s = 39.7 ft>s
Ans.
For the ball to travel from A to C, the initial and final horizontal positions are
(s0)x = 0 and sx = (21 + s) ft, respectively. The time is t = t2 = 0.6825 s.
+ B
A;
sx = (s0)x + (y0)x t
21 + s = 0 + 39.72(0.6825)
s = 6.11 ft
Ans.
87
7.5 ft
3 ft
C
A
21 ft
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12–110. It is observed that the skier leaves the ramp A at an
angle uA = 25° with the horizontal. If he strikes the ground
at B, determine his initial speed vA and the time of flight tAB.
A
+
:
B
4m
s = v0 t
4
100 a b = vA cos 25°tAB
5
A+cB
vA
uA
A
s = s0 + v0 t +
3
100 m
1
a t2
2 c
B
1
3
-4 - 100 a b = 0 + vA sin 25°tAB + (-9.81)t2AB
5
2
Solving,
vA = 19.4 m>s
Ans.
tAB = 4.54 s
Ans.
12–111. When designing a highway curve it is required
that cars traveling at a constant speed of 25 m>s must not
have an acceleration that exceeds 3 m>s2. Determine the
minimum radius of curvature of the curve.
Acceleration: Since the car is traveling with a constant speed, its tangential
component of acceleration is zero, i.e., at = 0. Thus,
a = an =
v2
r
252
r
r = 208 m
3 =
Ans.
*12–112. At a given instant, a car travels along a circular
curved road with a speed of 20 m>s while decreasing its speed
at the rate of 3 m>s2. If the magnitude of the car’s acceleration
is 5 m>s2, determine the radius of curvature of the road.
Acceleration: Here, the
at = -3 m>s2. Thus,
car’s
tangential
component
of
acceleration
of
a = 2at 2 + an 2
5 = 4A -3 B 2 + an 2
an = 4 m>s2
an =
v2
r
4 =
202
r
5
4
r = 100 m
Ans.
88
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•12–113. Determine the maximum constant speed a
race car can have if the acceleration of the car cannot
exceed 7.5 m>s2 while rounding a track having a radius of
curvature of 200 m.
Acceleration: Since the speed of the race car is constant, its tangential component of
acceleration is zero, i.e., at = 0. Thus,
a = an =
7.5 =
v2
r
v2
200
n = 38.7 m>s
Ans.
12–114. An automobile is traveling on a horizontal
circular curve having a radius of 800 ft. If the acceleration of
the automobile is 5 ft>s2, determine the constant speed at
which the automobile is traveling.
Acceleration: Since the automobile is traveling at a constant speed, at = 0.
y2
Thus, an = a = 5 ft>s2. Applying Eq. 12–20, an = , we have
r
y = 2ran = 2800(5) = 63.2 ft>s
Ans.
12–115. A car travels along a horizontal circular curved
road that has a radius of 600 m. If the speed is uniformly
increased at a rate of 2000 km>h2, determine the magnitude
of the acceleration at the instant the speed of the car is
60 km>h.
at = ¢
2
2000 km 1000 m
1h
ba
b = 0.1543 m>s2
ba
2
1 km
3600 s
h
y = a
60 km 1000 m
1h
ba
ba
b = 16.67 m>s
h
1 km
3600 s
an =
y2
16.672
= 0.4630 m>s2
=
r
600
a = 2a2t + a2n = 20.15432 + 0.46302 = 0.488 m>s2
Ans.
89
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*12–116. The automobile has a speed of 80 ft>s at point A
and an acceleration a having a magnitude of 10 ft>s2, acting
in the direction shown. Determine the radius of curvature
of the path at point A and the tangential component of
acceleration.
t
A
u ⫽ 30⬚
a
Acceleration: The tangential acceleration is
at = a cos 30° = 10 cos 30° = 8.66 ft>s2
Ans.
n
and the normal acceleration is an = a sin 30° = 10 sin 30° = 5.00 ft>s2. Applying
y2
Eq. 12–20, an = , we have
r
r =
y2
802
=
= 1280 ft
an
5.00
Ans.
•12–117. Starting from rest the motorboat travels around
the circular path, r = 50 m, at a speed v = (0.8t) m>s,
where t is in seconds. Determine the magnitudes of the
boat’s velocity and acceleration when it has traveled 20 m.
r ⫽ 50 m
Velocity: The time for which the boat to travel 20 m must be determined first.
ds = ydt
L0
20 m
v
t
0.8 tdt
L0
t = 7.071 s
ds =
The magnitude of the boat’s velocity is
y = 0.8 (7.071) = 5.657 m>s = 5.66 m>s
Ans.
Acceleration: The tangential accelerations is
#
at = y = 0.8 m>s2
To determine the normal acceleration, apply Eq. 12–20.
an =
5.6572
y2
=
= 0.640 m>s2
r
50
Thus, the magnitude of acceleration is
a = 2a2t + a2n = 20.82 + 0.6402 = 1.02 m>s2
Ans.
90
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12–118. Starting from rest, the motorboat travels around
the circular path, r = 50 m, at a speed v = (0.2t2) m>s,
where t is in seconds. Determine the magnitudes of the
boat’s velocity and acceleration at the instant t = 3 s.
r ⫽ 50 m
Velocity: When t = 3 s, the boat travels at a speed of
y = 0.2 A 32 B = 1. 80 m>s
Ans.
#
Acceleration: The tangential acceleration is at = y = (0.4t) m>s2. When t = 3 s,
at = 0.4 (3) = 1.20 m>s2
To determine the normal acceleration, apply Eq. 12–20.
an =
1.802
y2
=
= 0.0648 m>s2
r
50
Thus, the magnitude of acceleration is
a = 2a2t + a2n = 21.202 + 0.06482 = 1.20 m>s2
Ans.
12–119. A car moves along a circular track of radius 250 ft,
and its speed for a short period of time 0 … t … 2 s is
v = 3(t + t2) ft>s, where t is in seconds. Determine the
magnitude of the car’s acceleration when t = 2 s. How far
has it traveled in t = 2 s?
v = 3 A t + t2 B
at =
dv
= 3 + 6t
dt
When t = 2 s,
at = 3 + 6(2) = 15 ft>s2
an =
C 3(2 + 22) D 2
v2
=
= 1.296 ft>s2
r
250
a = 2(15)2 + (1.296)2 = 15.1 ft>s2
Ans.
ds = v dt
L
ds =
¢s =
L0
2
3 A t + t2 B dt
2
3 2
t + t3 d
2
0
¢s = 14 ft
Ans.
91
v
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*12–120. The car travels along the circular path such that
its speed is increased by at = (0.5et) m>s2, where t is in
seconds. Determine the magnitudes of its velocity and
acceleration after the car has traveled s = 18 m starting
from rest. Neglect the size of the car.
L0
y
dy =
L0
s ⫽ 18 m
t
0.5e t dt
y = 0.5(e t - 1)
L0
18
ds = 0.5
L0
t
(e t - 1)dt
ρ ⫽ 30 m
18 = 0.5(e - t - 1)
t
Solving,
t = 3.7064 s
y = 0.5(e 3.7064 - 1) = 19.85 m>s = 19.9 m>s
#
at = y = 0.5e t ƒ t = 3.7064 s = 20.35 m>s2
an =
Ans.
19.852
y2
=
= 13.14 m>s2
r
30
a = 2a2t + a2n = 220.352 + 13.142 = 24.2 m>s2
Ans.
y
•12–121. The train passes point B with a speed of 20 m>s
which is decreasing at at = – 0.5 m>s2. Determine the
magnitude of acceleration of the train at this point.
x
y ⫽ 200 e 1000
Radius of Curvature:
x
x
x
dy
1
= 200 a
be 1000 = 0.2e 1000
dx
1000
d2y
2
dx
r =
= 0.2a
2
x
400 m
x
x
1
.
be 1000 = 0.2 A 10-3 B e 1000
1000
c1 + a
dx2
C 1 + ¢ 0.2
2 3>2
dy
b d
dx
d2y
2
=
x
2
e 1000 ≤ S
` 0.2 A 10 B
-3
x
e 1000
`
3>2
6
= 3808.96 m
x = 400 m
Acceleration:
#
a t = v = -0.5 m>s2
an =
B
A
y = 200e 1000
202
v2
=
= 0.1050 m>s2
r
3808.96
The magnitude of the train’s acceleration at B is
a = 2a2t + a2n = 2 A -0.5 B 2 + 0.10502 = 0.511 m>s2
Ans.
92
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12–122. The train passes point A with a speed of 30 m>s
and begins to decrease its speed at a constant rate of
at = – 0.25 m>s2. Determine the magnitude of the
acceleration of the train when it reaches point B, where
sAB = 412 m.
y
x
y ⫽ 200 e 1000
B
A
x
400 m
Velocity: The speed of the train at B can be determined from
vB 2 = vA 2 + 2a t (sB - sA)
vB 2 = 302 + 2(-0.25)(412 - 0)
vB = 26.34 m>s
Radius of Curvature:
x
y = 200e1000
dy
x
= 0.2e1000
dx
d2y
dx2
= 0.2 A 10-3 B e1000
x
2
r =
B1+ a
2
2
dy
b R
dx
d2y
dx2
3>2
2
x
1000
C 1 + £ 0.2e
=
3>2
≥ S
x
2
2 0.2 A 10-3 B e1000
6
= 3808.96 m
x = 400 m
Acceleration:
#
a t = v = -0.25 m>s2
an =
v2
26.342
= 0.1822 m>s2
=
r
3808.96
The magnitude of the train’s acceleration at B is
a = 2a2t + a2n = 2 A -0.5 B 2 + 0.18222 = 0.309 m>s2
Ans.
93
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12–123. The car passes point A with a speed of 25 m>s
after which its speed is defined by v = (25 - 0.15s) m>s.
Determine the magnitude of the car’s acceleration when it
reaches point B, where s = 51.5 m.
y
y ⫽ 16 ⫺
B
16 m
Velocity: The speed of the car at B is
vB = C 25 - 0.15 A 51.5 B D = 17.28 m>s
Radius of Curvature:
y = 16 -
1 2
x
625
dy
= -3.2 A 10-3 B x
dx
d2y
dx2
= -3.2 A 10-3 B
r =
B1 + a
2
dy 2 3>2
b B
dx
d2y
dx2
c1 + a -3.2 A 10-3 B xb d
2 3>2
=
2
2 -3.2 A 10-3 B 2
4
= 324.58 m
x = 50 m
Acceleration:
vB 2
17.282
= 0.9194 m>s2
=
r
324.58
dv
at = v
= A 25 - 0.15s B A -0.15 B = A 0.225s - 3.75 B m>s2
ds
an =
When the car is at B A s = 51.5 m B
a t = C 0.225 A 51.5 B - 3.75 D = -2.591 m>s2
Thus, the magnitude of the car’s acceleration at B is
a = 2a2t + a2n = 2(-2.591)2 + 0.91942 = 2.75 m>s2
Ans.
94
1 2
x
625
s
A x
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*12–124. If the car passes point A with a speed of 20 m>s
and begins to increase its speed at a constant rate of
at = 0.5 m>s2, determine the magnitude of the car’s
acceleration when s = 100 m.
y
y ⫽ 16 ⫺
B
16 m
Velocity: The speed of the car at C is
vC 2 = vA 2 + 2a t (sC - sA)
vC 2 = 202 + 2(0.5)(100 - 0)
vC = 22.361 m>s
Radius of Curvature:
y = 16 -
1 2
x
625
dy
= -3.2 A 10-3 B x
dx
d2y
dx2
= -3.2 A 10-3 B
r =
B1 + a
2
dy 2 3>2
b R
dx
d2y
dx2
c1 + a -3.2 A 10-3 B xb d
2 3>2
=
2
冷-3.2 A 10-3 B 冷
4
= 312.5 m
x=0
Acceleration:
#
a t = v = 0.5 m>s
an =
vC 2
22.3612
= 1.60 m>s2
=
r
312.5
The magnitude of the car’s acceleration at C is
a = 2a2t + a2n = 20.52 + 1.602 = 1.68 m>s2
Ans.
95
1 2
x
625
s
A x
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•12–125. When the car reaches point A it has a speed of
25 m>s. If the brakes are applied, its speed is reduced by
1>2
at = ( - 41 t ) m>s2. Determine the magnitude of acceleration
of the car just before it reaches point C.
r ⫽ 250 m
C
B
30⬚
Velocity: Using the initial condition v = 25 m>s at t = 0 s,
dv = at dt
v
L25 m>s
t
1
- t1>2 dt
4
L0
dv =
v = a25 -
1 3>2
t b m>s
6
(1)
Position: Using the initial conditions s = 0 when t = 0 s,
L
L0
ds =
L
vdt
t
s
1
a25- t 3>2 bdt
6
L0
ds =
s = a25t -
1 5>2
t bm
15
p
Acceleration: When the car reaches C, sC = 200 + 250 a b = 330.90 m. Thus,
6
330.90 = 25t -
1 5>2
t
15
Solving by trial and error,
t = 15.942 s
Thus, using Eq. (1).
vC = 25 -
1
(15.942)3>2 = 14.391 m>s
6
1
#
(a t)C = v = - A 15.9421>2 B = -0.9982 m>s2
4
A an B C =
vC 2
14.3912
=
= 0.8284 m>s2
r
250
The magnitude of the car’s acceleration at C is
a = 4A at B C 2 + A an B C 2 = 4(-0.9982)2 + 0.82842 = 1.30 m>s2
96
Ans.
A
200 m
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12–126. When the car reaches point A, it has a speed of
25 m>s. If the brakes are applied, its speed is reduced by
at = (0.001s - 1) m>s2. Determine the magnitude of
acceleration of the car just before it reaches point C.
r ⫽ 250 m
C
B
A
Velocity: Using the initial condition v = 25 m>s at t = 0 s,
v
L25 m>s
200 m
30⬚
dv = ads
vdv =
L0
s
(0.001s - 1)ds
v = 20.001s2 - 2s + 625
p
Acceleration: When the car is at point C, sC = 200 + 250a b = 330.90 m. Thus,
6
the speed of the car at C is
vC = 40.001 A 330.902 B - 2(330.90) + 625 = 8.526 m>s2
#
(at)C = v = [0.001(330.90) - 1] = -0.6691 m>s2
(an)C =
vC 2
8.5262
= 0.2908 m>s2
=
r
250
The magnitude of the car’s acceleration at C is
a = 2(at)C 2 + (an)C 2 = 2(-0.6691)2 + 0.29082 = 0.730 m>s2
Ans.
12–127. Determine the magnitude of acceleration of the
airplane during the turn. It flies along the horizontal
circular path AB in 40 s, while maintaining a constant speed
of 300 ft>s.
A
Acceleration: From the geometry in Fig. a, 2f + 60° = 180° or f = 60°. Thus,
u
p
= 90° - 60° or u = 60° = rad.
2
3
sAB = vt = 300 A 40 B = 12 000 ft
Thus,
r =
sAB
12 000
36 000
=
=
ft
p
u
p>3
an =
v2
3002
= 7.854 ft>s2
=
r
36 000>p
Since the airplane travels along the circular path with a constant speed, at = 0. Thus,
the magnitude of the airplane’s acceleration is
a = 2at 2 + an 2 = 202 + 7.8542 = 7.85 ft>s2
Ans.
97
B
60⬚
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*12–128. The airplane flies along the horizontal circular path
AB in 60 s. If its speed at point A is 400 ft>s, which decreases at
a rate of at = (–0.1t) ft>s2, determine the magnitude of the
plane’s acceleration when it reaches point B.
A
B
60⬚
Velocity: Using the initial condition v = 400 ft>s when t = 0 s,
dv = at dt
v
L400 ft>s
L0
dv =
t
-0.1tdt
v = A 400 - 0.05t2 B ft>s
Position: Using the initial condition s = 0 when t = 0 s,
L
L0
ds =
s
ds =
L
L0
vdt
t
A 400 - 0.05t2 B dt
s = A 400t - 0.01667 t3 B ft
u
Acceleration: From the geometry, 2f + 60° = 180° or f = 60°. Thus, = 90° - 60°
2
p
or u = 60° = rad.
3
sAB = 400 A 60 B - 0.01667 A 603 B = 20 400 ft
r =
sAB
20 400
61200
=
=
ft
p
u
p>3
vB = 400 - 0.05 A 602 B = 220 ft>s
A an B B =
AatBB
vB 2
2202
= 2.485 ft>s2
=
r
61 200>p
#
= v = -0.1(60) = -6ft>s2
The magnitude of the airplane’s acceleration is
a = 2at 2 + an 2 = 2(-6)2 + 2.4852 = 6.49 ft>s2
Ans.
98
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•12–129. When the roller coaster is at B, it has a speed of
25 m>s, which is increasing at at = 3 m>s2. Determine the
magnitude of the acceleration of the roller coaster at this
instant and the direction angle it makes with the x axis.
y
y ⫽ 1 x2
100
A
s
B
x
Radius of Curvature:
1 2
x
100
dy
1
=
x
dx
50
y =
d2y
=
dx2
30 m
1
50
B1 + a
r =
2
dy 2 3>2
b R
dx
d2y
dx2
=
B1 + a
2
2 3>2
1
xb R
50
2 1 2
50
5
= 79.30 m
x = 30 m
Acceleration:
#
a t = v = 3 m>s2
an =
vB 2
252
= 7.881 m>s2
=
r
79.30
The magnitude of the roller coaster’s acceleration is
a = 2at 2 + an 2 = 232 + 7.8812 = 8.43 m>s2
Ans.
The angle that the tangent at B makes with the x axis is f = tan-1 ¢
dy
1
2
≤ = tan-1 c A 30 B d = 30.96°.
dx x = 30 m
50
As shown in Fig. a, an is always directed towards the center of curvature of the path. Here,
a = tan-1 a
an
7.881
b = tan-1 a
b = 69.16°. Thus, the angle u that the roller coaster’s acceleration makes
at
3
with the x axis is
u = a - f = 38.2°
Ans.
99
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12–130. If the roller coaster starts from rest at A and its
speed increases at at = (6 - 0.06 s) m>s2, determine the
magnitude of its acceleration when it reaches B where
sB = 40 m.
y
y ⫽ 1 x2
100
A
s
B
x
Velocity: Using the initial condition v = 0 at s = 0,
dv = at dt
L0
v
L0
vdv =
s
30 m
A 6 - 0.06s B ds
v = a 212s - 0.06s2 b m>s
Thus,
(1)
vB = 412 A 40 B - 0.06 A 40 B 2 = 19.60 m>s
Radius of Curvature:
1 2
x
100
dy
1
=
x
dx
50
y =
d2y
dx2
r =
=
1
50
B1 + a
2
dy 2 3>2
b R
dx
d2y
dx2
=
B1 + a
2
2 3>2
1
xb R
50
2 1 2
50
5
= 79.30 m
x = 30 m
Acceleration:
#
a t = v = 6 - 0.06(40) = 3.600 m>s2
an =
v2
19.602
= 4.842 m>s2
=
r
79.30
The magnitude of the roller coaster’s acceleration at B is
a = 2at 2 + an 2 = 23.6002 + 4.8422 = 6.03 m>s2
Ans.
100
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12–131. The car is traveling at a constant speed of 30 m>s.
The driver then applies the brakes at A and thereby reduces
the car’s speed at the rate of at = (-0.08v) m>s2, where v is
in m>s. Determine the acceleration of the car just before it
reaches point C on the circular curve. It takes 15 s for the
car to travel from A to C.
45⬚
A
100 m
Velocity: Using the initial condition v = 30 m>s when t = 0 s,
dv
a
dt =
L0
v
t
dt =
L30 m>s
t = 12.5 In
-
dv
0.08v
30
v
v = A 30e-0.08t B m>s
(1)
Position: Using the initial condition s = 0 when t = 0 s,
ds = vdt
L0
s
ds =
L0
t
30e-0.08t dt
s = C 375 A 1 - e-0.08t B D m
(2)
Acceleration:
sC = 375 A 1 - e -0.08(15) B = 262.05 m
sBC = sC - sB = 262.05 - 100 = 162.05 m
r =
sBC
162.05
=
= 206.33 m
u
p>4
vC = 30e -0.08(15) = 9.036 m>s
A an B C =
AatBC
C
B
vC 2
9.0362
= 0.3957 m>s2
=
r
206.33
#
= v = -0.08(9.036) = -0.7229 m>s2
The magnitude of the car’s acceleration at point C is
a = 2(at)C 2 + (an)C 2 = 2(-0.7229)2 + 0.39572 = 0.824 m>s2
101
Ans.
s
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*12–132. The car is traveling at a speed of 30 m>s. The
driver applies the brakes at A and thereby reduces the
speed at the rate of at = A - 18t B m>s2, where t is in seconds.
Determine the acceleration of the car just before it reaches
point C on the circular curve. It takes 15 s for the car to
travel from A to C.
45⬚
A
100 m
Velocity: Using the initial condition v = 30 m>s when t = 0 s,
dv = at dt
v
L30 m>s
t
1
- tdt
8
L0
dv =
v = a30 -
1 2
t b m>s
16
Position: Using the initial condition s = 0 when t = 0 s,
ds = vdt
L0
s
L0
ds =
s = a 30t -
t
a30 -
1 2
t bdt
16
1 3
t bm
48
Acceleration:
sC = 30(15) -
1
A 153 B = 379.6875 m
48
sBC = sC - sB = 379.6875 - 100 = 279.6875 m
r =
sBC
279.6875
=
= 356.11 m
u
p>4
vC = 30 -
1
A 152 B = 15.9375 m>s
16
A a t B C = v = - (15) = -1.875 m>s2
#
A an B C =
C
B
1
8
yC 2
15.93752
= 0.7133 m>s2
=
r
356.11
The magnitude of the car’s acceleration at point C is
a = 4A at B C 2 + A an B C 2 = 4A -1.875 B 2 + 0.71332 = 2.01 m>s2
102
Ans.
s
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•12–133. A particle is traveling along a circular curve
having a radius of 20 m. If it has an initial speed of 20 m>s
and then begins to decrease its speed at the rate of
at = ( -0.25s) m>s2, determine the magnitude of the
acceleration of the particle two seconds later.
Velocity: Using the initial condition v = 20 m>s at s = 0.
vdv = ads
v
L20 m>s
vdv =
L0
s
-0.25 sds
v = a 4400 - 0.25s2 b m>s
Position: Using the initial condition s = 0 when t = 0 s.
dt =
L0
ds
v
s
t
dt =
L0 2400 - 0.25s2
t = 2 sin-3 a
ds
s
= 2
L0 21600 - s2
ds
s
b
40
s = A 40 sin (t>2) B m
When t = 2 s,
s = 40 sin (2>2) = 33.659 m
Acceleration:
#
at = v = -0.25(33.659) = -8.415 m>s2
v = 4400 - 0.25 A 33.6592 B = 10.81 m>s
an =
v2
10.812
= 5.8385 m>s2
=
r
20
The magnitude of the particle’s acceleration is
a = 4a2t + a2n = 4A -8.415 B 2 + 5.83852 = 10.2 m>s2
103
Ans.
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12–134. A racing car travels with a constant speed of
240 km>h around the elliptical race track. Determine the
acceleration experienced by the driver at A.
y
y2 ⫽ 1
x2 ⫹ ––
––
16
4
B
Radius of Curvature:
x2
b
16
y2 = 4 a1 2y
A
2 km
dy
x
= dx
2
4 km
(1)
dy
x
= dx
4y
(2)
Differentiating Eq. (1),
2a
dy dy
d2 y
1
b a b + 2y 2 = dx dx
2
dx
-
2
dy
=
dx2
dy 2
1
- 2a b
2
dx
2y
dy 2
b
dx
T
4y
1 + 4a
2
dy
= -D
dx2
(3)
Substituting Eq. (2) into Eq. (3) yields
d2y
= -B
dx2
4y2 + x2
16y3
R
Thus,
r =
B1 + a
2
dy 2 3>2
b R
dx
d2y
2
dx
=
B 1 + a-
2
2
x 2 3>2
b R
4y
4y2 + x2
3
16y
2
¢1 +
=
¢
x2 3>2
≤
16y2
4y2 + x2
16y3
=
≤
A 16y2 + x2 B 3>2
4 A 4y2 + x2 B
Acceleration: Since the race car travels with a constant speed along the track, at = 0.
At x = 4 km and y = 0,
rA =
A 16y2 + x2 B 3>2
4 A 4y2 + x2 B
3
=
x = 4 km
y=0
A 0 + 42 B 3>2
4 A 0 + 42 B
= 1 km = 1000 m
The speed of the race car is
v = a 240
km 1000 m
1h
ba
ba
b = 66.67 m>s
h
1 km
3600 s
Thus,
aA =
v2
66.672
= 4.44 m>s2
=
rA
1000
Ans.
104
x
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12–135. The racing car travels with a constant speed of
240 km>h around the elliptical race track. Determine the
acceleration experienced by the driver at B.
y
x2 ⫹ ––
y2 ⫽ 1
––
16
4
B
Radius of Curvature:
A
2
x
b
16
y2 = 4a 1 -
2 km
4 km
dy
x
= 2y
dx
2
(1)
dy
x
= dx
4y
(2)
Differentiating Eq. (1),
2a
dy dy
d2y
1
b a b + 2y 2 = dx dx
2
dx
-
2
dy
=
dx2
dy 2
1
- 2a b
2
dx
2y
dy 2
b
dx
T
4y
1 + 4a
2
dy
= -D
dx2
(3)
Substituting Eq. (2) into Eq. (3) yields
d2y
= -B
dx2
4y2 + x2
16 y3
R
Thus,
r =
B1+ a
2
dy 2 3>2
b R
dx
d2y
2
dx
=
2
B1+ a -
2
x 2 3>2
b R
4y
4y2 + x2
3
16y
¢1 +
=
2
¢
x2 3>2
≤
16y2
4y2 + x2
3
16y
=
A 16y2 + x2 B 3>2
≤
4 A 4y2 + x2 B
Acceleration: Since the race car travels with a constant speed along the track, at = 0.
At x = 0 and y = 2 km
rB =
A 16y2 + x2 B 3>2
4 A 4y2 + x2 B
3
=
x=0
y = 2 km
C 16 A 22 B + 0 D 3>2
4 C 4 A 22 B + 0 D
= 8 km = 8000 m
The speed of the car is
v = a 240
km 1000 m
1h
ba
ba
b = 66.67 m>s
h
1 km
3600 s
Thus,
aB =
v2
66.672
= 0.556 m>s2
=
rB
8000
Ans.
105
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*12–136. The position of a particle is defined by
r = 52 sin (p4 )t i + 2 cos (p4 )t j + 3 t k6 m, where t is in
seconds. Determine the magnitudes of the velocity and
acceleration at any instant.
Velocity:
r = c2 sin a
v =
p
p
t bi + 2 cos a t b j + 3tk d m
4
4
dr
p
p
p
p
= c cos a t b i sin a t bj + 3k d m>s
dt
2
4
2
4
The magnitude of the velocity is
v = 2vx 2 + vy 2 + vz 2 =
=
Aa
p 2
p
p 2
p
cos tb + a- sin t b + 32
2
4
2
4
p2 + 9 = 3.39 m>s
A 4
Ans.
Acceleration:
a =
dv
p2
p
p2
p
= c - sin a tbi cos a t bj d m>s2
dt
8
4
8
4
Thus, the magnitude of the particle’s acceleration is
a = 2ax 2 + ay 2 + az 2 =
B
¢-
p2
p2
p 2
p2
p 2
m>s2 = 1.23 m>s2
sin t ≤ + ¢ cos t ≤ =
8
8
4
8
4
106
Ans.
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•12–137. The position of a particle is defined by
r = 5t 3 i + 3t 2 j + 8t k6 m, where t is in seconds. Determine
the magnitude of the velocity and acceleration and the
radius of curvature of the path when t = 2 s.
Velocity:
r = C t3i + 3t2j + 8tk D m
v =
dr
= C 3t2i + 6tj + 8k D m>s
dt
When t = 2 s,
v = c3 A 22 B i + 6 A 2 B j + 8k d = [12i + 12j + 8k] m>s
The magnitude of the velocity is
v = 2vx 2 + vy 2 + vz 2 = 2122 + 122 + 82 = 18.76 = 18.8 m>s
Ans.
Acceleration:
a =
dv
= [6ti + 6j] m>s2
dt
When t = 2 s,
a = C 6 A 2 B i + 6j D = [12i + 6j] m>s2
Thus, the magnitude of the particle’s acceleration is
a = 2ax 2 + ay 2 + az 2 = 2122 + 62 + 02 = 13.42 = 13.4 m>s2
Ans.
Since at is parallel to v, its magnitude can be obtained by the vector dot product
v
at = a # uv, where uv = = 0.6396i + 0.6396j + 0.4264k. Thus,
v
at = A 12i + 6j B
Thus,
# A 0.6396i + 0.6396j + 0.4264k B
= 11.51 m>s2
a = 2at 2 + an 2
13.42 = 211.512 + an 2
an = 6.889 m>s2
an =
v2
r
6.889 =
18.762
r
r = 51.1 m
Ans.
107
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12–138. Car B turns such that its speed is increased by
(at)B = (0.5et) m>s2, where t is in seconds. If the car starts
from rest when u = 0°, determine the magnitudes of its
velocity and acceleration when the arm AB rotates u = 30°.
Neglect the size of the car.
B
5m
A
dy
Velocity: The speed y in terms of time t can be obtained by applying a =
.
dt
dy = adt
L0
y
L0
dy =
t
0.5et dt
y = 0.5 A et - 1 B
[1]
30°
pb = 2.618 m.
180°
The time required for the car to travel this distance can be obtained by applying
When u = 30°, the car has traveled a distance of s = ru = 5a
y =
ds
.
dt
ds = ydt
L0
2.618 m
ds =
L0
t
0.5 A e - 1 B dt
t
2.618 = 0.5 A et - t - 1 B
Solving by trial and error
t = 2.1234 s
Substituting t = 2.1234 s into Eq. [1] yields
y = 0.5 A e2.1234 - 1 B = 3.680 m>s = 3.68 m>s
Ans.
Acceleration: The tangential acceleration for the car at t = 2.1234 s is
at = 0.5e2.1234 = 4.180 m>s2. To determine the normal acceleration, apply Eq. 12–20.
an =
3.6802
y2
=
= 2.708 m>s2
r
5
The magnitude of the acceleration is
a = 2a2t + a2n = 24.1802 + 2.7082 = 4.98 m>s2
Ans.
108
u
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12–139. Car B turns such that its speed is increased by
(at)B = (0.5et) m>s2, where t is in seconds. If the car starts
from rest when u = 0°, determine the magnitudes of its
velocity and acceleration when t = 2 s. Neglect the size of
the car.
B
Velocity: The speed y in terms of time t can be obtained by applying a =
dy
.
dt
5m
dy = adt
L0
When t = 2 s,
v
dy =
L0
t
t
A
0.5e dt
u
y = 0.5 A et - 1 B
y = 0.5 A e2 - 1 B = 3.195 m>s = 3.19 m>s
Ans.
Acceleration: The tangential acceleration of the car at t = 2 s is
at = 0.5e2 = 3.695 m>s2. To determine the normal acceleration, apply Eq. 12–20.
an =
3.1952
y2
=
= 2.041 m>s2
r
5
The magnitude of the acceleration is
a = 2a2t + a2n = 23.6952 + 2.0412 = 4.22 m>s2
Ans.
*12–140. The truck travels at a speed of 4 m>s along a
circular road that has a radius of 50 m. For a short distance
from s = 0, its speed is then increased by at = (0.05s) m>s2,
where s is in meters. Determine its speed and the magnitude
of its acceleration when it has moved s = 10 m.
50 m
Velocity: The speed y in terms of position s can be obtained by applying ydy = ads.
ydy = ads
y
L4 m>s
y =
At s = 10 m,
ydy =
L0
s
0.05sds
A 20.05s2 + 16 B m>s
y = 20.05(102) + 16 = 4.583 m>s = 4.58 m>s
Ans.
Acceleration: The tangential acceleration of the truck at s = 10 m is
at = 0.05 (10) = 0.500 m>s2. To determine the normal acceleration, apply Eq. 12–20.
an =
y2
4.5832
= 0.420 m>s2
=
r
50
The magnitude of the acceleration is
a = 2a2t + a2n = 20.5002 + 0.4202 = 0.653 m>s2
Ans.
109
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•12–141. The truck travels along a circular road that has a
radius of 50 m at a speed of 4 m>s. For a short distance when
t = 0, its speed is then increased by at = (0.4t) m>s2, where
t is in seconds. Determine the speed and the magnitude of
the truck’s acceleration when t = 4 s.
50 m
Velocity: The speed y in terms of time t can be obtained by applying a =
dy
.
dt
dy = adt
y
L4 m>s
dy =
L0
t
0.4tdt
y = A 0.2t2 + 4 B m>s
y = 0.2 A 42 B + 4 = 7.20 m>s
When t = 4 s,
Ans.
Acceleration: The tangential acceleration of the truck when t = 4 s is
at = 0.4(4) = 1.60 m>s2. To determine the normal acceleration, apply Eq. 12–20.
an =
y2
7.202
= 1.037 m>s2
=
r
50
The magnitude of the acceleration is
a = 2a2t + a2n = 21.602 + 1.0372 = 1.91 m>s2
Ans.
110
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12–142. Two cyclists, A and B, are traveling
counterclockwise around a circular track at a constant speed
of 8 ft>s at the instant shown. If the speed of A is increased
at (at)A = (sA) ft>s2, where sA is in feet, determine the
distance measured counterclockwise along the track from B
to A between the cyclists when t = 1 s. What is the
magnitude of the acceleration of each cyclist at this instant?
sA
A
B
r ⫽ 50 ft
Distance Traveled: Initially, the distance between the cyclists is d0 = ru
= 50 a
120°
pb = 104.72 ft. When t = 1 s, cyclist B travels a distance of sB = 8(1)
180°
= 8 ft. The distance traveled by cyclist A can be obtained as follows
yAdyA = aA dsA
yA
L8 ft>s
yA dyA =
L0
sA
sA dsA
yA = 2s2A + 64
dt =
1s
L0
L0
[1]
dsA
yA
sA
dsA
2s2A + 64
sA
1 = sinh-1 a b
8
dt =
sA = 9.402 ft
Thus, the distance between the two cyclists after t = 1 s is
d = d0 + sA - sB = 104.72 + 9.402 - 8 = 106 ft
Ans.
Acceleration: The tangential acceleration for cyclist A and B at t = 1 s is
(at)A = sA = 9.402 ft>s2 and (at)B = 0 (cyclist B travels at constant speed),
respectively. At t = 1 s, from Eq. [1], yA = 29.4022 + 64 = 12.34 ft>s. To
determine normal acceleration, apply Eq. 12–20.
(an)A =
y2A
12.342
=
= 3.048 ft>s2
r
50
(an)B =
y2B
82
= 1.28 ft>s2
=
r
50
sB
u ⫽ 120⬚
The magnitude of the acceleration for cyclist A and B are
aA = 2(at)2A + (an)2A = 29.4022 + 3.0482 = 9.88 ft>s2
aB = 2(at)2B + (an)2B = 202 + 1.282 = 1.28 ft>s2
Ans.
Ans.
111
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12–143. A toboggan is traveling down along a curve which
can be approximated by the parabola y = 0.01x2.
Determine the magnitude of its acceleration when it
reaches point A, where its speed is vA = 10 m>s, and it is
increasing at the rate of (at)A = 3 m>s2.
y
y ⫽ 0.01x2
A
36 m
Acceleration: The radius of curvature of the path at point A must be determined
d2y
dy
first. Here,
= 0.02x and 2 = 0.02, then
dx
dx
r =
[1 + (dy>dx)2]3>2
2
2
|d y>dx |
=
x
60 m
[1 + (0.02x)2]3>2
2
= 190.57 m
|0.02|
x = 60 m
To determine the normal acceleration, apply Eq. 12–20.
102
y2
= 0.5247 m>s2
=
r
190.57
an =
#
Here, at = yA = 3 m>s. Thus, the magnitude of accleration is
a = 2a2t + a2n = 232 + 0.52472 = 3.05 m>s2
Ans.
*12–144. The jet plane is traveling with a speed of 120 m>s
which is decreasing at 40 m>s2 when it reaches point A.
Determine the magnitude of its acceleration when it is at this
point. Also, specify the direction of flight, measured from the
x axis.
y
x)
y ⫽ 15 ln ( ––
80
80 m
A
x
y = 15 lna b
80
dy
15 2
= 0.1875
=
x x = 80 m
dx
d2y
dx2
= -
15 2
= -0.002344
x2 x = 80 m
r4
=
dy 2 3>2
) D
C 1 + (dx
`
x = 80 m
=
an =
d2y
2
dx
`
4
x = 80 m
[1 + (0.1875)2]3>2
= 449.4 m
|-0.002344|
(120)2
v2
=
= 32.04 m>s2
r
449.4
at = -40 m>s2
a = 2( -40)2 + (32.04)2 = 51.3 m>s2
Ans.
Since
dy
= tan u = 0.1875
dx
u = 10.6°
Ans.
112
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•12–145. The jet plane is traveling with a constant speed
of 110 m>s along the curved path. Determine the magnitude
of the acceleration of the plane at the instant it reaches
point A (y = 0).
y = 15 lna
y
x)
y ⫽ 15 ln ( ––
80
x
b
80
80 m
x
A
dy
15 2
= 0.1875
=
x x = 80 m
dx
d2y
dx2
= -
r4
15 2
= -0.002344
x2 x = 80 m
=
x = 80 m
dy 2 3>2
) D
C 1 + (dx
d2y
`
2
dx
`
4
x = 80 m
C 1 + (0.1875) D
2 3>2
=
an =
= 449.4 m
| -0.002344|
(110)2
y2
= 26.9 m>s2
=
r
449.4
Since the plane travels with a constant speed, at = 0. Hence
a = an = 26.9 m>s2
Ans.
12–146. The motorcyclist travels along the curve at a
constant speed of 30 ft>s. Determine his acceleration when
he is located at point A. Neglect the size of the motorcycle
and rider for the calculation.
y
500
dy
500
= -0.05
= - 2 2
dx
x x = 100 ft
d2y
2
dx
=
y ⫽ –—
x
1000 2
= 0.001
x3 x = 100 ft
r4
=
x = 100 ft
dy 2 3>2
)]
[1 + (dx
`
d 2y
dx
2
`
100 ft
4
x = 100 ft
2 3>2
=
an =
[1 + (-0.05) ]
|0.001|
= 1003.8 ft
302
y2
= 0.897 ft>s2
=
r
1003.8
Since the motorcyclist travels with a constant speed, at = 0. Hence
a = an = 0.897 ft>s2
Ans.
113
v ⫽ 30 ft/s
A
x
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12–147. The box of negligible size is sliding down along a
curved path defined by the parabola y = 0.4x2. When it is at
A (xA = 2 m, yA = 1.6 m), the speed is vB = 8 m>s and the
increase in speed is dvB>dt = 4 m>s2. Determine the
magnitude of the acceleration of the box at this instant.
y
A
2
y ⫽ 0.4x
y = 0.4 x2
x
dy
2
= 0.8x 2
= 1.6
dx x = 2 m
x=2 m
d2y
dx2
2
2m
= 0.8
x=2 m
dy 2 3>2
) D
C 1 + (dx
r =
`
d 2y
2
dx
`
4
=
C 1 + (1.6)2 D 3>2
|0.8|
= 8.396 m
x=2 m
2
an =
yB
82
= 7.622 m>s2
=
r
8.396
a = 2a2t + a2n = 2(4)2 + (7.622)2 = 8.61 m>s2
Ans.
*12–148. A spiral transition curve is used on railroads to
connect a straight portion of the track with a curved
portion. If the spiral is defined by the equation
y = (10 - 6)x3, where x and y are in feet, determine the
magnitude of the acceleration of a train engine moving with
a constant speed of 40 ft>s when it is at point x = 600 ft.
y
y ⫽ (10⫺6)x3
y = (10)-6x3
v ⫽ 40 ft/s
dy
2
= 3(10)-6x2 2
= 1.08
dx x = 600 ft
x = 600 ft
d2y
dx2
2
= 6(10)-6x 2
x = 600 ft
r4
=
x = 600 ft
an =
x
= 3.6(10)-3
x = 600 ft
dy 2 3>2
[1 + (dx
) D
`
2
dy
2
dx
`
4
600 ft
=
[1 + (1.08)2]3>2
|3.6(10)-3|
= 885.7 ft
x = 600 ft
y2
402
= 1.81 ft>s2
=
r
885.7
a = 2a2t + a2n = 20 + (1.81)2 = 1.81 ft>s2
Ans.
114
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•12–149. Particles A and B are traveling counterclockwise around a circular track at a constant speed of
8 m>s. If at the instant shown the speed of A begins to
increase by (at)A =(0.4sA) m>s2, where sA is in meters,
determine the distance measured counterclockwise along
the track from B to A when t = 1 s. What is the magnitude
of the acceleration of each particle at this instant?
A
sA
u ⫽ 120⬚
sB
B
r⫽5m
Distance Traveled: Initially the distance between the particles is
d0 = rdu = 5a
120°
bp = 10.47 m
180°
When t = 1 s, B travels a distance of
dB = 8(1) = 8 m
The distance traveled by particle A is determined as follows:
vdv = ads
v
L8 m>s
L0
vdv =
s
0.4 sds
v = 0.6325 2s2 + 160
ds
v
dt =
L0
s
t
dt =
(1)
L0 0.6325 2s2 + 160
ds
1
2s2 + 160 + s
£ InC
S≥
0.6325
2160
1 =
s = 8.544 m
Thus the distance between the two cyclists after t = 1 s is
d = 10.47 + 8.544 - 8 = 11.0 m
Ans.
Acceleration:
For A, when t = 1 s,
A a t B A = vA = 0.4 A 8.544 B = 3.4176 m>s2
#
vA = 0.632528.5442 + 160 = 9.655 m>s
(a n)A =
v2A
9.6552
= 18.64 m>s2
=
r
5
The magnitude of the A’s acceleration is
aA = 23.41762 + 18.642 = 19.0 m>s2
Ans.
For B, when t = 1 s,
A at B B = vA = 0
#
(an)B =
v2B
82
= 12.80 m>s2
=
r
5
The magnitude of the B’s acceleration is
aB = 202 + 12.802 = 12.8 m>s2
Ans.
115
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12–150. Particles A and B are traveling around a circular
track at a speed of 8 m>s at the instant shown. If the speed of B
is increasing by (at)B = 4 m>s2, and at the same instant A has
an increase in speed of (at)A = 0.8t m>s2, determine how long
it takes for a collision to occur. What is the magnitude of the
acceleration of each particle just before the collision occurs?
A
sA
u ⫽ 120⬚
sB
B
Distance Traveled: Initially the distance between the two particles is d0 = ru
= 5a
120°
pb = 10.47 m. Since particle B travels with a constant acceleration,
180°
distance can be obtained by applying equation
sB = (s0)B + (y0)B t +
sB = 0 + 8t +
1
a t2
2 c
1
(4) t2 = A 8 t + 2 t2 B m
2
[1]
The distance traveled by particle A can be obtained as follows.
dyA = aA dt
yA
L8 m>s
dyA =
L0
t
0.8 tdt
yA = A 0.4 t2 + 8 B m>s
[2]
dsA = yA dt
L0
sA
dsA =
L0
t
A 0.4 t2 + 8 B dt
sA = 0.1333t3 + 8 t
In order for the collision to occur
sA + d0 = sB
0.1333t3 + 8t + 10.47 = 8 t + 2 t2
Solving by trial and error
t = 2.5074 s = 2.51 s
Ans.
240°
pb + sB. This equation will result in
Note: If particle A strikes B then, sA = 5a
180°
t = 14.6 s 7 2.51 s.
Acceleration: The tangential acceleration for particle A and B when t = 2.5074 are
(a t)A = 0.8 t = 0.8 (2.5074) = 2.006 m>s2 and (a t)B = 4 m>s2, respectively. When
t = 2.5074 s, from Eq. [1], yA = 0.4 A 2.50742 B + 8 = 10.51 m>s and yB = (y0)B + ac t
= 8 + 4(2.5074) = 18.03 m>s.To determine the normal acceleration, apply Eq. 12–20.
(an)A =
y2A
10.512
= 22.11 m>s2
=
r
5
(an)B =
y2B
18.032
= 65.01 m>s2
=
r
5
The magnitude of the acceleration for particles A and B just before collision are
aA = 2(at)2A + (an)2A = 22.0062 + 22.112 = 22.2 m>s2
aB = 2(at)2B + (an)2B = 242 + 65.012 = 65.1 m>s2
Ans.
Ans.
116
r⫽5m
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12–151. The race car travels around the circular track with
a speed of 16 m>s. When it reaches point A it increases its
speed at at = (43 v1>4) m>s2, where v is in m>s. Determine the
magnitudes of the velocity and acceleration of the car when
it reaches point B. Also, how much time is required for it to
travel from A to B?
y
A
200 m
B
4 1
v4
3
at =
dv = at dt
dv =
4 1
v 4 dt
3
v
L16
1
dv
0.75
=
1
4
v
L0
t
dt
v
v4冷16 = t
3
v4 - 8 = t
4
v = (t + 8)3
ds = v dt
L0
s
ds =
L0
t
4
(t + 8)3 dt
t
2
3
s = (t + 8)3 2
7
0
s =
2
3
(t + 8)3 - 54.86
7
For s =
7
3
p
(200) = 100p = (t + 8)3 - 54.86
2
7
t = 10.108 s = 10.1 s
Ans.
4
3
v = (10.108 + 8) = 47.551 = 47.6 m>s
at =
an =
Ans.
1
4
(47.551)4 = 3.501 m>s2
3
(47.551)2
v2
= 11.305 m>s2
=
r
200
a = 2(3.501)2 + (11.305)2 = 11.8 m>s2
Ans.
117
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*12–152. A particle travels along the path y = a + bx
+ cx2, where a, b, c are constants. If the speed of the particle
is constant, v = v0, determine the x and y components
of velocity and the normal component of acceleration
when x = 0.
y = a + bx + cx2
#
#
#
y = bx + 2 c x x
$
$
#
$
y = b x + 2 c (x)2 + 2 c xx
#
#
y = bx
When x = 0,
#
#
v20 + x 2 + b2 x 2
#
vx = x =
v0
21 + b2
v0 b
Ans.
21 + b2
v20
an =
r
vy =
Ans.
dy 2 2
C 1 + A dx
B D
3
r =
`
d2y
2
dx
`
dy
= b + 2c x
dx
d2y
dx2
= 2c
At x = 0,
an =
r =
(1 + b2)3>2
2c
2 c v20
Ans.
(1 + b2)3>2
118
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•12–153. The ball is kicked with an initial speed
vA = 8 m>s at an angle uA = 40° with the horizontal. Find
the equation of the path, y = f(x), and then determine the
normal and tangential components of its acceleration when
t = 0.25 s.
y
vA = 8 m/s
uA ⫽ 40⬚
A
x
Horizontal Motion: The horizontal component of velocity is (v0)x = 8 cos 40°
= 6.128 m>s and the initial horizontal and final positions are (s0)x = 0 and sx = x,
respectively.
+ B
A:
sx = (s0)x + (y0)x t
x = 0 + 6.128t
[1]
Vertical Motion: The vertical component of initial velocity is (y0)y = 8 sin 40°
= 5.143 m>s. The initial and final vertical positions are (s0)y = 0 and sy = y,
respectively.
A+cB
sy = (s0)y + (y0)y t +
y = 0 + 5.143t +
1
(a ) t2
2 cy
1
(-9.81) A t2 B
2
[2]
Eliminate t from Eqs [1] and [2], we have
y = {0.8391x - 0.1306x2} m = {0.839x - 0.131x2} m
The tangent of the path makes an angle u = tan-1
Ans.
3.644
= 42.33° with the x axis.
4
Acceleration: When t = 0.25 s, from Eq. [1], x = 0 + 6.128(0.25) = 1.532 m. Here,
dy
dy
= 0.8391 - 0.2612x. At x = 1.532 m,
= 0.8391 - 0.2612(1.532) = 0.4389
dx
dx
and the tangent of the path makes an angle u = tan-1 0.4389 = 23.70° with the x axis.
The magnitude of the acceleration is a = 9.81 m>s2 and is directed downward. From
the figure, a = 23.70°. Therefore,
at = a sin a = 9.81 sin 23.70° = 3.94 m>s2
Ans.
an = a cos a = 9.81 cos 23.70° = 8.98 m>s2
Ans.
119
y
x
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12–154. The motion of a particle is defined by the
equations x = (2t + t2) m and y = (t2) m, where t is in
seconds. Determine the normal and tangential components
of the particle’s velocity and acceleration when t = 2 s.
Velocity: Here, r =
E A 2 t + t 2 B i + t2 j F m.To determine the velocity v, apply Eq. 12–7.
v =
dr
= {(2 + 2t) i + 2tj } m>s
dt
When t = 2 s, v = [2 + 2(2)]i + 2(2)j = {6i + 4j} m>s. Then y = 262 + 42
= 7.21 m>s. Since the velocity is always directed tangent to the path,
yn = 0
and
yt = 7.21 m>s
The velocity v makes an angle u = tan-1
Ans.
4
= 33.69° with the x axis.
6
Acceleration: To determine the acceleration a, apply Eq. 12–9.
a =
Then
dv
= {2i + 2j} m>s2
dt
a = 222 + 22 = 2.828 m>s2
The acceleration a makes an angle f = tan-1
figure, a = 45° - 33.69 = 11.31°. Therefore,
2
= 45.0° with the x axis. From the
2
an = a sin a = 2.828 sin 11.31° = 0.555 m>s2
Ans.
at = a cos a = 2.828 cos 11.31° = 2.77 m>s2
Ans.
120
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12–155. The motorcycle travels along the elliptical track
at a constant speed v. Determine the greatest magnitude of
the acceleration if a 7 b.
y
b
x 2 y2
⫹
⫽1
a2 b2
a
b
2a2 - x2, we have
a
Acceleration: Differentiating twice the expression y =
dy
bx
= dx
a2a2 - x2
d2y
2
dx
= -
ab
2
(a - x2)3>2
The radius of curvature of the path is
r =
dy 2 3>2
c1 + a b d
dx
2
2
dy
dx2
B1 + ¢ =
2
2-
2
bx
2
2
ab
2
a2a - x
(a2 - x2)3>2
3>2
≤ R
=
c1 +
3>2
b2x2
d
2
2
a (a - x )
ab
2
[1]
(a2 - x2)3>2
To have the maximum normal acceleration, the radius of curvature of the path must be
a minimum. By observation, this happens when y = 0 and x = a. When x : a,
3>2
3>2
b2x2
b2x2
b3x3
b2x2
7 7 7 1. Then, c1 + 2 2
d :c 2 2
d
.
= 3 2
2
2
2
2
a (a - x )
a (a - x )
a (a - x )
a (a - x2)3>2
2
Substituting this value into Eq. [1] yields r =
r =
b2 3
x . At x = a,
a4
b2 3
b2
a
=
A
B
a
a4
To determine the normal acceleration, apply Eq. 12–20.
(an)max =
a
y2
y2
= 2 = 2 y2
r
b >a
b
Since the motorcycle is traveling at a constant speed, at = 0. Thus,
amax = (an)max =
a 2
y
b2
Ans.
121
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*12–156. A particle moves along# a circular path of radius
300 mm. If its angular velocity is u = (2t2) rad>s, where t is
in seconds, determine the magnitude of the particle’s
acceleration when t = 2 s.
Time Derivatives:
#
$
r = r = 0
#
u = 2t2冷t = 2 s = 8 rad>s
$
u = 4t冷t = 2 s = 8 rad>s2
Velocity: The radial and transverse components of the particle’s velocity are
#
#
vr = r = 0
vu = ru = 0.3(8) = 2.4 m>s
Thus, the magnitude of the particle’s velocity is
v = 2vr 2 + vu 2 = 202 + 2.42 = 2.4 m>s
Ans.
Acceleration:
#
$
a r = r - ru2 = 0 - 0.3 A 82 B = -19.2 m>s2
$
##
a u = ru + 2ru = 0.3(8) + 0 = 2.4 m>s2
Thus, the magnitude of the particle’s acceleration is
a = 2ar 2 + au 2 = 2(-19.2)2 + 2.42 = 19.3 m>s2
Ans.
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•12–157. A particle moves along# a circular path of radius
300 mm. If its angular velocity is u = (3t2) rad>s, where t is
in seconds, determine the magnitudes of the particle’s
velocity and acceleration when u = 45°. The particle starts
from rest when u = 0°.
Time Derivatives: Using the initial condition u = 0° when t = 0 s,
du = 3t2dt
L0
u
du =
L0
t
3t2 dt
u = A t3 B rad
At u = 45° =
p
rad,
4
p
= t3
t = 0.9226 s
4
#
$
r = r = 0
#
u = 3t2冷t = 0.9226 s = 2.554 rad>s
$
u = 6t冷t = 0.9226 s = 5.536 rad>s2
Velocity:
#
vr = r = 0
#
vu = ru = 0.3(2.554) = 0.7661 m>s
Thus, the magnitude of the particle’s velocity is
v = 2vr 2 + vu 2 = 202 + 0.76612 = 0.766 m>s
Ans.
Acceleration:
#
$
a r = r - ru2 = 0 - 0.3 A 2.5542 B = -1.957 m>s2
$
##
a u = ru + 2ru = 0.3(5.536) + 0 = 1.661 m>s2
Thus, the magnitude of the particle’s acceleration is
a = 2ar 2 + au 2 = 2(-1.957)2 + 1.6612 = 2.57 m>s2
123
Ans.
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12–158. A particle moves along a circular path of radius
5 ft. If its position is u = (e0.5t) rad, where t is in seconds,
determine the magnitude of the particle’s acceleration
when u = 90°.
Time Derivative: At u = 90° =
p
= e0.5t
2
p
rad,
2
t = 0.9032 s
Using the result of t, the value of the first and second time derivative of r and u are
#
$
r = r = 0
#
u = 0.5e 0.5 t冷t = 0.9032 s = 0.7854 rad>s
$
u = 0.25e 0.5t冷t = 0.9032 s = 0.3927 rad>s2
Acceleration:
#
$
a r = r - ru2 = 0 - 5 A 0.78542 B = -3.084 ft>s2
$
##
a u = ru + 2ru = 5(0.3927) + 0 = 1.963 ft>s2
Thus, the magnitude of the particle’s acceleration is
a = 2ar 2 + au 2 = 2(-3.084)2 + 1.9632 = 3.66 ft>s2
124
Ans.
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12–159. The position of a particle is described by
r = (t3 + 4t - 4) mm and u = (t3>2) rad, where t is in
seconds. Determine the magnitudes of the particle’s
velocity and acceleration at the instant t = 2 s.
Time Derivatives: The first and second time derivative of r and u when t = 2 s are
r = t3 + 4t - 4冷t = 2 s = 12 m
u = t3>2
#
r = 3t2 + 4冷t = 2 s = 16 m>s
#
3
u = t1>2 2
= 2.121 rad>s
2
t=2 s
$
r = 6t冷t = 2 s = 12 m>s2
$
3
u = t-1>2 2
= 0.5303 rad>s2
4
t=2 s
Velocity:
#
vr = r = 16 m>s
#
vu = ru = 12(2.121) = 25.46 m>s
Thus, the magnitude of the particle’s velocity is
v = 2vr 2 + vu 2 = 2162 + 25.462 = 30.1 m>s
Ans.
Acceleration:
#
$
a r = r - ru2 = 12 - 12 A 2.1212 B = -42.0 m>s2
$
##
a u = ru + 2ru = 12(0.5303) + 2(16)(2.121) = 74.25 m>s2
Thus, the magnitude of the particle’s acceleration is
a = 2ar 2 + au 2 = 2(-42.0)2 + 74.252 = 85.3 m>s2
Ans.
125
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*12–160. The position of a particle is described by
r = (300e–0.5t) mm and u = (0.3t2) rad, where t is in seconds.
Determine the magnitudes of the particle’s velocity and
acceleration at the instant t = 1.5 s.
Time Derivatives: The first and second time derivative of r and u when t = 1.5 s are
r = 300e-0.5t冷t - 1.5 s = 141.77 mm
#
r = -150e-0.5t冷t-1.5 s = -70.85 mm>s
$
r = 75e-0.5t 冷t-1.5 s = 35.43 mm>s2
u = 0.3t2 rad
#
u = 0.6t冷t - 1.5 s = 0.9 rad>s
$
u = 0.6 rad>s2
Velocity:
#
vr = r = -70.85 mm>s
#
vu = ru = 141.71(0.9) = 127.54 mm>s
Thus, the magnitude of the particle’s velocity is
v = 2vr 2 + vu 2 = 2(-70.85)2 + 127.542 = 146 mm>s
Ans.
Acceleration:
#
$
ar = r - ru2 = 35.43 - 141.71 A 0.92 B = -79.36 mm>s2
$
##
a u = ru + 2ru = 141.71(0.6) + 2(-70.85)(0.9) = -42.51 mm>s2
Thus, the magnitude of the particle’s acceleration is
a = 2ar 2 + au 2 = 2(-79.36)2 + (-42.51)2 = 90.0 mm>s2
Ans.
•12–161. An airplane is flying in a straight line with a
velocity of 200 mi>h and an acceleration of 3 mi>h2. If the
propeller has a diameter of 6 ft and is rotating at an
angular rate of 120 rad>s, determine the magnitudes of
velocity and acceleration of a particle located on the tip
of the propeller.
yPl = ¢
200 mi 5280 ft
1h
≤¢
≤¢
≤ = 293.3 ft>s
h
1 mi
3600 s
aPl = ¢
3 mi 5280 ft
1h 2
≤
¢
≤
¢
≤ = 0.001 22 ft>s2
1 mi
3600 s
h2
yPr = 120(3) = 360 ft>s
y = 2y2Pl + y2Pr = 2(293.3)2 + (360)2 = 464 ft>s
aPr =
y2Pr
r
Ans.
2
=
(360)
= 43 200 ft>s2
3
a = 2a2Pl + a2Pr = 2(0.001 22)2 + (43 200)2 = 43.2(103) ft>s2
126
Ans.
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12–162. A particle moves along a circular path having a
radius of 4 in. such that its position as a function of time is
given by u = (cos 2t) rad, where t is in seconds. Determine
the magnitude of the acceleration of the particle when
u = 30°.
When u =
p
6
rad,
p
6
= cos 2t
t = 0.5099 s
#
du
= -1.7039 rad>s
u =
= -2 sin 2t 2
dt
t = 0.5099 s
$
d2u
= -2.0944 rad>s2
u = 2 = -4 cos 2t 2
dt
t = 0.5099 s
r = 4
#
r = 0
$
r = 0
#
$
ar = r - ru2 = 0 - 4(-1.7039)2 = -11.6135 in.>s2
$
##
au = ru + 2ru = 4(-2.0944) + 0 = -8.3776 in.>s2
a = 2a2r + a2u = 2(-11.6135)2 + (-8.3776)2 = 14.3 in.>s2
Ans.
12–163. A particle travels around a limaçon, defined by
the equation r = b - a cos u, where a and b are constants.
Determine the particle’s radial and transverse components
of velocity and acceleration as a function of u and its time
derivatives.
r = b - a cos u
#
#
r = a sin uu
#
$
#
r = a cos uu2 + a sin uu
#
#
vr = r = a sin uu
Ans.
#
vu = r u = (b - a cos u)u
#
#
$
#
$
ar = r - r u2 = a cos uu2 + a sin uu - (b - a cos u)u2
#
$
= (2a cos u - b) u2 + a sin u u
$
$
# #
# #
au = ru + 2 r u = (b - a cos u)u + 2aa sin uu bu
Ans.
Ans.
$
#
= (b - a cos u)u + 2au2 sin u
Ans.
127
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*12–164. A particle travels around a lituus, defined by the
equation r2u = a2, where a is a constant. Determine the
particle’s radial and transverse components of velocity and
acceleration as a function of u and its time derivatives.
r2u = a2
1
r = au- 2
3 #
1
r = aa - bu- 2u.
2
3$
3 3#
1
$
r = - a a - u - 2 u2 + u- 2 u b
2
2
1 -3 #
#
vr = r = - au 2 u
2
#
1 #
vu = ru = au- 2 u
Ans.
Ans.
#
3$
1 #
1
3 3#
$
ar = r - r u2 = - aa - u - 2 u2 + u- 2 u b - au- 2 u2
2
2
1 #
1 3$
3
= ac a u -2 - 1 bu- 2 u2 - u- 2 u d
4
2
Ans.
#
$
$
#
$
u2 - 1
1 -3 # #
- 12
au = ru + 2ru = a u u + 2(a)a - bu 2u au b = a B u - R u 2
2
u
Ans.
•12–165. A car travels along the circular curve of radius
r = # 300 ft. At the instant shown, its angular rate of rotation
is u = 0.4 rad>s, which is increasing at the rate of
$
u = 0.2 rad>s2. Determine the magnitudes of the car’s
velocity and acceleration at this instant.
A
r ⫽ 300 ft
.
Velocity: Applying Eq. 12–25, we have
#
#
yr = r = 0
yu = ru = 300(0.4) = 120 ft>s
u ⫽ 0.4 rad/s
u ⫽ 0.2 rad/s2
..
u
Thus, the magnitude of the velocity of the car is
y = 2y2r + y2u = 202 + 1202 = 120 ft>s
Ans.
Acceleration: Applying Eq. 12–29, we have
#
$
ar = r - ru2 = 0 - 300 A 0.42 B = -48.0 ft>s2
$
##
au = ru + 2ru = 300(0.2) + 2(0)(0.4) = 60.0 ft>s2
Thus, the magnitude of the acceleration of the car is
a = 2a2r + a2u = 2(-48.0)2 + 60.02 = 76.8 ft>s2
Ans.
128
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12–166. The slotted arm OA rotates counterclockwise
#
about O with a constant angular velocity of u. The motion of
pin B is constrained such that it moves on the fixed circular
surface and along the slot in OA. Determine the magnitudes
of the velocity and acceleration of pin B as a function of u.
B
r ⫽ 2 a cos u
u
O
Time Derivatives:
r = 2a cos u
#
#
r = -2a sin uu
##
$
#
$
$
r = -2a C cos uuu + sin uu D = -2a C cos uu2 + sin uu D
#
$
Since u is constant, u = 0. Thus,
#
$
r = -2a cos uu2
Velocity:
#
#
vr = r = -2a sin uu
#
#
vu = ru = 2a cos uu
Thus, the magnitude of the pin’s velocity is
# 2
# 2
2
2
v = 4vr + vu = 4A -2a sin uu B + A 2a cos uu B
#
#
= 44a2u 2 A sin2 u + cos2 u B = 2au
Ans.
Acceleration:
#
#
#
#
$
ar = r - ru2 = -2a cos uu2 - 2a cos uu2 = -4a cos uu2
$
# #
#
# #
au = ru + 2r u = 0 + 2 A -2a sin uu B u = -4a sin uu2
Thus, the magnitude of the pin’s acceleration is
#
a = 2ar 2 + au 2 = 4A -4a cos uu2 B 2 +
#
#
= 416a 2u4 A cos2 u + sin 2 u B = 4au2
A -4a sin uu2 B 2
#
Ans.
129
a
A
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12–167. The slotted arm OA rotates counterclockwise
about O such that when# u = p>4, arm OA is rotating with
$
an angular velocity of u and an angular acceleration of u.
Determine the magnitudes of the velocity and acceleration
of pin B at this instant. The motion of pin B is constrained
such that it moves on the fixed circular surface and along
the slot in OA.
B
r ⫽ 2 a cos u
u
O
Time Derivatives:
r = 2a cos u
#
#
r = -2a sin uu
##
#
#
#
$
r = -2a C cos uuu + sin uu D = -2a C cos uu2 + sin uu D
When u =
Velocity:
p
rad,
4
1
22
≤ = 22a
r冷u =
p
4
= 2a ¢
#
r冷u =
p
4
= -2a ¢
22
$
r冷u =
p
4
= -2a ¢
#
$
1 #2
1 $
u +
u ≤ = - 22a A u2 + u B
22
22
1
#
#
vr = r = - 22au
≤ u = - 22au
#
#
#
#
vu = ru = 22au
Thus, the magnitude of the pin’s velocity is
#
v = 2vr 2 + vu 2 = 4A - 22au B 2 +
A 22au B 2 = 2au
#
#
Ans.
Acceleration:
#
#
$
#
#
$
$
ar = r - ru2 = - 22a A u2 + u B - 22au2 = - 22a A 2u2 + u B
$
$
# #
$
#
# #
au = ru + 2r u = 22au + 2 A - 22au B A u B = 22a A u - 2u2 B
Thus, the magnitude of the pin’s acceleration is
#
$
$
#
a = 2ar 2 + au 2 = 4C - 22a A 2u2 + u B D 2 + C 22a A u - 2u2 B D 2
#
$
= 2a 44u4 + u2
130
Ans.
a
A
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*12–168. The car travels along the circular curve having a
radius r = 400
ft. At the instant shown, its angular rate of
#
rotation
is
u
=
0.025
rad>s, which is decreasing at the rate
$
u = -0.008 rad>s2. Determine the radial and transverse
components of the car’s velocity and acceleration at this
instant and sketch these components on the curve.
r = 400
#
u = 0.025
#
r = 0
r ⫽ 400 ft
.
u
$
r = 0
u = -0.008
#
vr = r = 0
#
vu = ru = 400(0.025) = 10 ft>s
#
ar = r - r u2 = 0 - 400(0.025)2 = -0.25 ft>s2
Ans.
Ans.
Ans.
au = r u + 2 r u = 400(-0.008) + 0 = -3.20 ft>s2
Ans.
•12–169. The car travels along the circular curve of radius
r = 400 ft with a constant speed
# of v = 30 ft>s. Determine
the angular rate of rotation u of the radial line r and the
magnitude of the car’s acceleration.
r ⫽ 400 ft
r = 400 ft
vr = r = 0
#
r = 0
$
r = 0
.
u
vu = r u = 400a ub
# 2
v = 6(0)2 + a400 u b = 30
u = 0.075 rad>s
$
u = 0
#
$
ar = r - r u2 = 0 - 400(0.075)2 = -2.25 ft>s2
Ans.
#
au = r u + 2 r u = 400(0) + 2(0)(0.075) = 0
a = 2(-2.25)2 + (0)2 = 2.25 ft>s2
Ans.
131
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12–170. Starting from rest, the boy runs outward in the
radial direction from the center of the platform with a
2
constant acceleration
# of 0.5 m>s . If the platform is rotating
at a constant rate u = 0.2 rad>s, determine the radial and
transverse components of the velocity and acceleration of
the boy when t = 3 s. Neglect his size.
u ⫽ 0.2 rad/s
0.5 m/s2
r
u
Velocity: When t = 3 s, the position of the boy is given by
s = (s0)r + (y0)r t +
r = 0 + 0 +
1
(a ) t2
2 cr
1
(0.5) A 32 B = 2.25 m
2
The boy’s radial component of velocity is given by
yr = (y0)r + (ac)r t
= 0 + 0.5(3) = 1.50 m>s
Ans.
The boy’s transverse component of velocity is given by
#
yu = ru = 2.25(0.2) = 0.450 m>s
Ans.
$
#
$
Acceleration: When t = 3 s, r = 2.25 m, r = yr = 1.50 m>s, r = 0.5 m>s2, u = 0.
Applying Eq. 12–29, we have
#
$
Ans.
ar = r - ru2 = 0.5 - 2.25 A 0.22 B = 0.410 m>s2
$
#
au = ru + 2ru = 2.25(0) + 2(1.50)(0.2) = 0.600 m>s2
Ans.
12–171. The small washer slides down the cord OA. When it
is at the midpoint, its speed is 200 mm>s and its acceleration
is 10 mm>s2. Express the velocity and acceleration of the
washer at this point in terms of its cylindrical components.
z
A
v, a
OA = 2(400)2 + (300)2 + (700)2 = 860.23 mm
700 mm
OB = 2(400)2 + (300)2 = 500 mm
z
O
u
500
b = 116 mm>s
vr = (200)a
860.23
x
700
b = 163 mm>s
860.23
Thus, v = {-116ur - 163uz} mm>s
ar = 10 a
Ans.
500
b = 5.81
860.23
au = 0
az = 10 a
y
300 mm
vu = 0
vz = (200)a
r
700
b = 8.14
860.23
Thus, a = {-5.81ur - 8.14uz} mm>s2
Ans.
132
400 mm
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*12–172. If arm OA rotates# counterclockwise with a
constant angular velocity of u = 2 rad>s, determine the
magnitudes of the velocity and acceleration of peg P at
u = 30°. The peg moves in the fixed groove defined by the
lemniscate, and along the slot in the arm.
r2 ⫽ (4 sin 2 u)m2
r
O
Time Derivatives:
r2 = 4 sin 2u
#
#
2rr = 8 cos 2uu
#
4 cos 2uu
#
r = B
R m>s
r
#
u = 2 rad>s
#
$
$
#
2 A rr + r2 B = 8 A -2 sin 2uu2 + cos 2u u B
$
#
#
4 A cos 2uu - 2 sin 2uu2 B - r2
$
S m>s2
r = C
r
$
u = 0
At u = 30°,
r冷u = 30° = 24 sin 60° = 1.861 m
(4 cos 60°)(2)
#
r冷u = 30° =
= 2.149 m>s
1.861
4 C 0 - 2 sin 60° A 22 B D - (2.149)2
$
r冷u = 30° =
= -17.37 m>s2
1.861
Velocity:
#
vr = r = 2.149 m>s
#
vu = ru = 1.861(2) = 3.722 m>s
Thus, the magnitude of the peg’s velocity is
v = 2vr 2 + vu 2 = 22.1492 + 3.7222 = 4.30 m>s
Ans.
Acceleration:
#
$
a r = r - ru2 = -17.37 - 1.861 A 22 B = -24.82 m>s2
$
# #
a u = r u + 2r u = 0 + 2(2.149)(2) = 8.597 m>s2
Thus, the magnitude of the peg’s acceleration is
a = 2ar 2 + au 2 = 2(-24.82)2 + 8.5972 = 26.3 m>s2
133
Ans.
u
P
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•12–173. The peg moves in the curved slot defined by the
lemniscate, and through
the slot in the arm. At u = 30°, the
#
angular
velocity
is
u
=
2
rad>s, and the angular acceleration
$
is u = 1.5 rad>s2. Determine the magnitudes of the velocity
and acceleration of peg P at this instant.
r2 ⫽ (4 sin 2 u)m2
r
O
Time Derivatives:
#
#
2rr = 8 cos 2uu
#
4 cos 2uu
#
r = ¢
≤ m>s
r
#
u = 2 rad>s
$
#
$
#
2 a rr + r2 b = 8 a -2 sin 2uu + cos 2uu2 b
$
#
#
4 A cos 2uu - 2 sin 2uu2 B - r2
$
S m>s2
r = C
r
$
u = 1.5 rad>s2
At u = 30°,
r冷u = 30° = 24 sin 60° = 1.861 m
(4 cos 60°)(2)
#
r冷u = 30° =
= 2.149 m>s
1.861
4 C cos 60°(1.5) - 2 sin 60° A 22 B D - (2.149)2
$
r冷u = 30° =
= -15.76 m>s2
1.861
Velocity:
#
vr = r = 2.149 m>s
#
vu = ru = 1.861(2) = 3.722 m>s
Thus, the magnitude of the peg’s velocity is
v = 2ar 2 + au 2 = 22.1492 + 3.7222 = 4.30 m>s
Ans.
Acceleration:
#
$
a r = r - r u2 = -15.76 - 1.861 A 22 B = -23.20 m>s2
$
# #
au = ru + 2r u = 1.861(1.5) + 2(2.149)(2) = 11.39 m>s2
Thus, the magnitude of the peg’s acceleration is
a = 2ar 2 + au 2 = 2(-23.20)2 + 11.392 = 25.8 m>s2
134
Ans.
u
P
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12–174. The airplane on the amusement park ride moves
along a path defined by the equations r = 4 m,
u = (0.2t) rad, and z = (0.5 cos u) m, where t is in seconds.
Determine the cylindrical components of the velocity and
acceleration of the airplane when t = 6 s.
r = 4m
#
r = 0
$
r = 0
z = 0.5 cos u
z
u = 0.2t冷t = 6 s = 1.2 rad
#
u = 0.2 rad>s
$
u = 0
#
#
z = -0.5 sin u u冷u = 1.2 rad = -0.0932 m>s
#
$
$
z = -0.5 C cos uu2 + sin uu D 冷u = 1.2 rad = -0.007247 m>s2
#
yr = r = 0
#
yu = ru = 4(0.2) = 0.8 m>s
Ans.
Ans.
#
yz = z = -0.0932 m>s
#
$
a r = r - r u2 = 0 - 4(0.2)2 = -0.16 m>s2
$
# #
au = r u + 2r u = 4(0) + 2(0)(0.2) = 0
Ans.
Ans.
Ans.
$
az = z = -0.00725 m>s2
Ans.
135
u
r⫽4m
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12–175. The motion of peg P is constrained by the
lemniscate curved slot in OB and by the slotted arm OA. If
OA rotates # counterclockwise with a constant angular
velocity of u = 3 rad>s, determine the magnitudes of the
velocity and acceleration of peg P at u = 30°.
r
O
Time Derivatives:
r2 = 4 cos 2u
#
#
2rr = -8 sin 2uu
#
-4 sin 2uu
#
r = ¢
≤ m>s
r
#
u = 3 rad>s
$
#
$
#
2 A rr + r2 B = -8 A sin 2uu + 2 cos 2u u2 B
$
#
#
-4 A sin 2uu + 2 cos 2uu2 B - r2
$
r = C
S m>s2
r
$
u = 0
#
u = 3 rad>s. Thus, when u = 30°,
r冷u = 30° = 24 cos 60° = 22 m
-4 sin 60°(3)
#
r冷u = 30° =
= -7.348 m>s
22
-4 C 0 + 2 cos 60°(3)2 D - (-7.348)2
$
r冷u = 30° =
= -63.64 m>s2
22
Velocity:
#
vr = r = -7.348 m>s
A
P
#
vu = ru = 22(3) = 4.243 m>s
Thus, the magnitude of the peg’s velocity is
v = 2vr 2 + vu 2 = 27.3482 + (-4.243)2 = 8.49 m>s
Ans.
Acceleration:
#
$
ar = r - ru2 = -63.64 - 22(3)2 = -76.37 m>s2
$
##
au = ru + 2ru = 0 + 2( -7.348)(3) = -44.09 m>s2
Thus, the magnitude of the peg’s acceleration is
a = 2ar 2 + au 2 = 2(-76.37)2 + ( -44.09)2 = 88.2 m>s2
136
Ans.
u
r2 ⫽ (4 cos 2 u)m2
B
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*12–176. The motion of peg P is constrained by the
lemniscate curved slot in OB and by the slotted arm OA.
If
# OA rotates counterclockwise with an angular velocity of
u = (3t3/2) rad>s, where t is in seconds, determine the
magnitudes of the velocity and acceleration of peg P at
u = 30°. When t = 0, u = 0°.
r
O
Time Derivatives:
r2 = 4 cos 2u
#
#
2rr = -8 sin 2uu
#
-4 sin 2uu
#
r = ¢
≤ m>s
r
$
#
$
#
2 A rr + r2 B = -8 A sin 2uu + 2 cos 2u u2 B
$
#
#
-4 A sin 2uu + 2 cos 2uu2 B - r 2
$
r = C
S m>s2
r
#
du
= u = 3t3>2
dt
u
L0°
L0
du =
t
3t3>2dt
6
u = ¢ t5>2 ≤ rad
5
At u = 30° =
p
rad,
6
p
6
= t5>2
6
5
#
u = 3t3>2 2
t = 0.7177 s
= 1.824 rad>s
t = 0.7177 s
$
9
= 3.812 rad>s2
u = t1>2 2
2
t = 0.7177 s
Thus,
A
P
r冷u = 30° = 24 cos 60° = 22 m
-4 sin 60°(1.824)
#
r冷u = 30° =
= -4.468 m>s
22
-4 C sin 60°(3.812) + 2 cos 60°(1.824)2 D - (-4.468)2
$
= -32.86 m>s2
r冷u = 30° =
22
137
u
r2 ⫽ (4 cos 2 u)m2
B
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12–176. Continued
Velocity:
#
vr = r = -4.468 m>s
#
vu = ru = 22(1.824) = 2.579 m>s
Thus, the magnitude of the peg’s velocity is
v = 2vr 2 + vu 2 = 2(-4.468)2 + 2.5792 = 5.16 m>s
Ans.
Acceleration:
#
$
ar = r - r u2 = -32.86 - 22(1.824)2 = -37.57 m>s2
$
##
au = ru + 2ru = 22(3.812) + 2(-4.468)(1.824) = -10.91 m>s2
Thus, the magnitude of the peg’s acceleration is
a = 2ar 2 + au 2 = 2(-37.57)2 + (-10.91)2 = 39.1 m>s2
138
Ans.
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•12–177. The driver of the car maintains a constant speed
of 40 m>s. Determine the angular velocity of the camera
tracking the car when u = 15°.
r ⫽ (100 cos 2u) m
u
Time Derivatives:
r = 100 cos 2u
#
#
r = (-200 sin 2uu) m>s
At u = 15°,
r冷u = 15° = 100 cos 30° = 86.60 m
#
#
#
r冷u = 15° = -200 sin 30°u = -100u m>s
Velocity: Referring to Fig. a, vr = -40 cos f and vu = 40 sin f.
#
vr = r
#
-40 cos f = -100u
(1)
and
#
vu = ru
#
40 sin f = 86.60u
(2)
Solving Eqs. (1) and (2) yields
f = 40.89°
#
u = 0.3024 rad>s = 0.302 rad>s
Ans.
139
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12–178. When u = 15°, the car has a speed of 50 m>s
which is increasing at 6 m>s2. Determine the angular
velocity of the camera tracking the car at this instant.
Time Derivatives:
r ⫽ (100 cos 2u) m
r = 100 cos 2u
#
#
r = (-200 sin 2uu) m>s
$
#
$
r = -200 C sin 2uu + 2 cos 2uu2 D m>s2
u
At u = 15°,
r冷u = 15° = 100 cos 30° = 86.60 m
#
#
#
r冷u = 15° = -200 sin 30°u = -100u m>s
$
#
#
r冷u = 15° = -200 C sin 30° u + 2 cos 30° u2 D =
A -100u - 346.41u2 B m>s2
$
#
Velocity: Referring to Fig. a, vr = -50 cos f and vu = 50 sin f. Thus,
#
vr = r
#
-50 cos f = -100u
and
(1)
#
vu = ru
#
50 sin f = 86.60u
(2)
Solving Eqs. (1) and (2) yields
f = 40.89°
#
u = 0.378 rad>s
Ans.
12–179. If the cam
# rotates clockwise with a constant
angular velocity of u = 5 rad>s, determine the magnitudes
of the velocity and acceleration of the follower rod AB at
the instant u = 30°. The surface of the cam has a shape of
limaçon defined by r = (200 + 100 cos u) mm.
r ⫽ (200 ⫹ 100 cos u) mm
u
Time Derivatives:
r = (200 + 100 cos u) mm
#
#
r = (-100 sin uu) mm>s
$
#
$
r = -100 C sin uu + cos uu2 D mm>s2
#
u = 5 rad>s
$
u = 0
When u = 30°,
r冷u = 30° = 200 + 100 cos 30° = 286.60 mm
#
r冷u = 30° = -100 sin 30°(5) = -250 mm>s
#
r冷u = 30° = -100 C 0 + cos 30° A 52 B D = -2165.06 mm>s2
Velocity: The radial component gives the rod’s velocity.
#
vr = r = -250 mm>s
Ans.
Acceleration: The radial component gives the rod’s acceleration.
#
$
ar = r - ru2 = -2156.06 - 286.60(52) = -9330 mm>s2
Ans.
140
A
B
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*12–180. At the instant u = # 30°, the cam rotates with a
clockwise angular
$ velocity of u = 5 rad>s and and angular
acceleration of u = 6 rad>s2. Determine the magnitudes of
the velocity and acceleration of the follower rod AB at this
instant. The surface of the cam has a shape of a limaçon
defined by r = (200 + 100 cos u) mm.
r ⫽ (200 ⫹ 100 cos u) mm
u
Time Derivatives:
r = (200 - 100 cos u) mm
#
#
r = (-100 sin uu) mm>s
$
$
#
r = -100 C sin uu + cos uu2 D mm>s2
When u = 30°,
r冷u = 30° = 200 + 100 cos 30° = 286.60 mm
#
r冷u = 30° = -100 sin 30°(5) = -250 mm>s
$
r冷u = 30° = -100 C sin 30°(6) + cos 30° A 52 B D = -2465.06 mm>s2
Velocity: The radial component gives the rod’s velocity.
#
vr = r = -250 mm>s
Ans.
Acceleration: The radial component gives the rod’s acceleration.
#
$
ar = r - ru2 = -2465.06 - 286.60 A 52 B = -9630 mm>s2
141
Ans.
A
B
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•12–181. The automobile travels from a parking deck
down along a cylindrical spiral ramp at a constant speed of
v = 1.5 m>s. If the ramp descends a distance of 12 m for
every full revolution, u = 2p rad, determine the magnitude
of the car’s acceleration as it moves along the ramp,
r = 10 m. Hint: For part of the solution, note that the
tangent to the ramp at any point is at an angle of
f = tan - 1 (12>32p(10)4) = 10.81° from the horizontal. Use
this to determine the velocity # components vu and vz, which
#
in turn are used to determine u and z.
f = tan-1 a
10 m
12 m
12
b = 10.81°
2p(10)
v = 1.5 m>s
vr = 0
vu = 1.5 cos 10.81° = 1.473 m>s
vz = -1.5 sin 10.81° = -0.2814 m>s
Since
r = 10
#
r = 0
#
vu = r u = 1.473
r = 0
u =
1.473
= 0.1473
10
Since u = 0
#
ar = r - r u2 = 0 - 10(0.1473)2 = -0.217
#$
au = r u + 2 r u = 10(0) + 2(0)(0.1473) = 0
$
az = z = 0
a = 2(-0.217)2 + (0)2 + (0)2 = 0.217 m>s2
Ans.
142
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12–182. The box slides down the helical ramp with a
constant speed of v = 2 m>s. Determine the magnitude of
its acceleration. The ramp descends a vertical distance of
1 m for every full revolution. The mean radius of the ramp is
r = 0.5 m.
Velocity: The inclination angle of the ramp is f = tan-1
0.5 m
1m
L
1
= tan-1 B
R = 17.66°.
2pr
2p(0.5)
Thus, from Fig. a, vu = 2 cos 17.66° = 1.906 m>s and vz = 2 sin 17.66° = 0.6066 m>s. Thus,
#
vu = ru
#
1.906 = 0.5u
#
u = 3.812 rad>s
#
$
#
$
Acceleration: Since r = 0.5 m is constant, r = r = 0. Also, u is constant, then u = 0.
Using the above results,
#
$
ar = r - r u2 = 0 - 0.5(3.812)2 = -7.264 m>s2
$
##
au = r u + 2ru = 0.5(0) + 2(0)(3.812) = 0
Since vz is constant az = 0. Thus, the magnitude of the box’s acceleration is
a = 2ar 2 + au 2 + az 2 = 2(-7.264)2 + 02 + 02 = 7.26 m>s2
143
Ans.
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12–183. The box slides down the helical ramp which is
defined by r = 0.5 m, u = (0.5t3) rad, and z = (2 – 0.2t2) m,
where t is in seconds. Determine the magnitudes of the
velocity and acceleration of the box at the instant
u = 2p rad.
0.5 m
1m
Time Derivatives:
r = 0.5 m
#
$
r = r = 0
#
u = A 1.5t2 B rad>s
$
u = 3(3t) rad>s2
z = 2 - 0.2t2
#
z = (-0.4t) m>s
$
z = -0.4 m>s2
When u = 2p rad,
2p = 0.5t3
t = 2.325 s
Thus,
#
u冷t = 2.325 s = 1.5(2.325)2 = 8.108 rad>s
$
u冷t = 2.325 s = 3(2.325) = 6.975 rad>s2
#
z冷t = 2.325 s = -0.4(2.325) = -0.92996 m>s
$
z冷t = 2.325 s = -0.4 m>s2
Velocity:
#
vr = r = 0
#
vu = ru = 0.5(8.108) = 4.05385 m>s
#
vz = z = -0.92996 m>s
Thus, the magnitude of the box’s velocity is
v = 2vr 2 + vu 2 + vz 2 = 202 + 4.053852 + (-0.92996)2 = 4.16 m>s
Ans.
Acceleration:
#
$
a r = r - r u2 = 0 - 0.5(8.108)2 = -32.867 m>s2
$
# #
a u = r u + 2r u = 0.5(6.975) + 2(0)(8.108)2 = 3.487 m>s2
$
az = z = -0.4 m>s2
Thus, the magnitude of the box’s acceleration is
a = 2ar 2 + au 2 + az 2 = 2(-32.867)2 + 3.4872 + (-0.4)2 = 33.1 m>s2 Ans.
144
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*12–184. Rod OA rotates
counterclockwise with a constant
#
angular velocity of u = 6 rad>s. Through mechanical means
#
collar B moves along the rod with a speed of r = (4t2) m>s,
where t is in seconds. If r = 0 when t = 0, determine the
magnitudes of velocity and acceleration of the collar when
t = 0.75 s.
A
r
u
O
Time Derivatives: Using the initial condition r = 0 when t = 0 s,
dr
#
= r = 4t2
dt
L0
r
dr =
L0
t
4t2 dt
4
r = c t3 d m
3
r =
4 3
t 2
= 0.5625 m
3 t = 0.75 s
#
r = 4t2 2
#
u = 6 rad>s
= 2.25 m>s
t = 0.75 s
$
r = 8t 2
$
u = 0
= 6 m>s2
t = 0.75 s
Velocity:
#
vr = r = 2.25 m>s
#
vu = ru = 0.5625(6) = 3.375 m>s
Thus, the magnitude of the collar’s velocity is
v = 2vr 2 + vu 2 = 22.252 + 3.3752 = 4.06 m>s
Ans.
Acceleration:
#
$
a r = r - r u2 = 6 - 0.5625 A 62 B = -14.25 m>s2
$
# #
a u = ru + 2r u = 0.5625(0) + 2(2.25)(6) = 27 m>s2
Thus, the magnitude of the collar’s acceleration is
a = 2ar 2 + au 2 = 2(-14.25)2 + (-27)2 = 30.5 m>s2
145
Ans.
B
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•12–185. Rod OA
# is rotating counterclockwise with an
angular velocity of u = (2t2) rad>s.Through mechanical means
#
collar B moves along the rod with a speed of r = (4t2) m>s. If
u = 0 and r = 0 when t = 0, determine the magnitudes of
velocity and acceleration of the collar at u = 60°.
A
r
Position: Using the initial condition u = 0 when t = 0 s,
u
O
#
du
= u = 2t2
dt
u
L0°
du =
L0
t
2t2dt
2
u = c t3 d rad
3
p
rad,
3
p
2
= t3
3
3
At u = 60° =
t = 1.162 s
Using the initial condition r = 0 when t = 0 s,
dr
#
= r = 4t2
dt
L0
r
dr =
L0
t
4t2 dt
4
r = c t3 d m
3
When t = 1.162 s au =
r =
p
radb ,
3
4 3
t 2
= 2.094 m
3 t = 1.162 s
Time Derivatives:
#
r = 4t2 2
#
u = 2t2 2
= 5.405 m>s
t = 1.162 s
$
r = 8t 2
= 2.703 rad>s
t = 1.162 s
$
u = 4t 2
= 9.300 m>s2
t = 1.162 s
Velocity:
#
vr = r = 5.405 m>s
= 4.650 rad>s2
t = 1.162 s
#
vu = ru = 2.094(2.703) = 5.660 m>s
Thus, the magnitude of the collar’s velocity is
v = 2vr 2 + vu 2 = 25.4052 + 5.6602 = 7.83 m>s
Ans.
Acceleration:
#
$
a r = r - r u2 = 9.300 - 2.094 A 2.7032 B = -5.998 m>s2
$
##
au = ru + 2ru = 2.094(4.650) + 2(5.405)(2.703) = 38.95 m>s2
Thus, the magnitude of the collar’s acceleration is
a = 2ar 2 + au 2 = 2(-5.998)2 + 38.952 = 39.4 m>s2
146
Ans.
B
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12–186. The slotted arm AB drives pin C through the spiral
groove described by the
equation r = a u. If the angular
#
velocity is constant at u, determine the radial and transverse
components of velocity and acceleration of the pin.
B
C
#
$
Time Derivatives: Since u is constant, then u = 0.
#
$
#
$
r = au
r = au = 0
r = au
r
u
Velocity: Applying Eq. 12–25, we have
#
#
yr = r = au
#
#
yu = ru = auu
Ans.
Acceleration: Applying Eq. 12–29, we have
#
#
#
$
ar = r - ru2 = 0 - auu2 = -auu2
$
# #
#
# #
au = ru + 2r u = 0 + 2(au)(u) = 2au2
Ans.
Ans.
12–187. The slotted arm AB drives pin C through the
spiral groove described by the equation r = (1.5 u) ft, where
u is in radians. If the arm starts from
# rest when u = 60° and
is driven at an angular velocity of u = (4t) rad>s, where t is
in seconds, determine the radial and transverse components
of velocity and acceleration of the pin C when t = 1 s.
B
C
r
#
$
Time Derivatives: Here, u = 4 t and u = 4 rad>s2.
#
$
#
$
r = 1.5 u = 1.5(4t) = 6t
r = 1.5 u = 1.5 (4) = 6 ft>s2
r = 1.5 u
Velocity: Integrate the angular rate,
L
u
du =
p
3
A
Ans.
L0
t
4tdt, we have u =
u
1
(6t2 + p) rad.
3
1
1
#
Then, r = b (6t2 + p) r ft. At t = 1 s, r = C 6(12) + p D = 4.571 ft, r = 6(1) = 6.00 ft>s
2
2
#
and u = 4(1) = 4 rad>s. Applying Eq. 12–25, we have
#
yr = r = 6.00 ft>s
#
yu = ru = 4.571 (4) = 18.3 ft>s
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
#
$
ar = r - ru2 = 6 - 4.571 A 42 B = -67.1 ft>s2
$
# #
au = r u + 2ru = 4.571(4) + 2(6) (4) = 66.3 ft>s2
Ans.
Ans.
147
A
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*12–188. The partial surface of the cam is that of a
logarithmic spiral r = (40e0.05u) mm, where u is in
# radians. If
the cam rotates at a constant angular velocity of u = 4 rad>s,
determine the magnitudes of the velocity and acceleration of
the point on the cam that contacts the follower rod at the
instant u = 30°.
u
r = 40e0.05 u
#
#
r = 2e 0.05u u
·
u ⫽ 4 rad/s
# 2
$
$
r = 0.1e 0.0 5 u au b + 2e 0.05 u u
u =
p
6
#
u = 4
$
u = 0
r = 40e 0.05A 6 B = 41.0610
p
p
#
r = 2e 0.05A 6 B (4) = 8.2122
p
$
r = 0.1e 0.05A 6 B (4)2 + 0 = 1.64244
#
vr = r = 8.2122
#
vu = ru = 41.0610(4) = 164.24
v = 2(8.2122)2 + (164.24)2 = 164 mm>s
#
$
ar = r - r u2 = 1.642 44 - 41.0610(4)2 = -655.33
$
# #
au = r u + 2 r u = 0 + 2(8.2122)(4) = 65.6976
Ans.
a = 2(-655.33)2 + (65.6976)2 = 659 mm>s2
Ans.
148
r ⫽ 40e0.05u
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•12–189. Solve $Prob. 12–188, if the cam has an angular
acceleration of u = 2 rad>s2 when its angular velocity is
#
u = 4 rad>s at u = 30°.
u
r = 40e 0.05u
#
r = 2e 0.05 u u
·
u ⫽ 4 rad/s
2
#
r = 0.1e 0.05 u aub + 2e 0.05 uu
u =
p
6
u = 4
u = 2
r = 40e 0.05 A 6 B = 41.0610
p
r = 2e 0.05 A 6 B (4) = 8.2122
p
p
p
$
r = 0.1e 0.05A 6 B (4)2 + 2e 0.05A 6 B (2) = 5.749
vr = r = 8.2122
#
vu = r u = 41.0610(4) = 164.24
v = 2(8.2122)2 + (164.24)2 = 164 mm>s
#
$
ar = r - r u2 = 5.749 - 41.0610(4)2 = -651.2
$
##
au = r u + 2ru = 41.0610(2) + 2(8.2122)(4) = 147.8197
Ans.
a = 2(-651.2)2 + (147.8197)2 = 668 mm>s2
Ans.
149
r ⫽ 40e0.05u
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12–190. A particle moves along an Archimedean
spiral
#
r = (8u) ft, where u is given in radians. If u = 4 rad>s
(constant), determine the radial and transverse components
of the particle’s velocity and acceleration at the instant
u = p>2 rad. Sketch the curve and show the components on
the curve.
y
r ⫽ (8 u) ft
r
#
$
Time Derivatives: Since u is constant, u = 0.
p
r = 8u = 8 a b = 4pft
2
u
#
#
r = 8 u = 8(4) = 32.0 ft>s
$
$
r = 8u = 0
x
Velocity: Applying Eq. 12–25, we have
#
y r = r = 32.0 ft>s
#
yu = ru = 4p (4) = 50.3 ft>s
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
#
$
ar = r - ru2 = 0 - 4p A 42 B = -201 ft>s2
$
# #
au = ru + 2r u = 0 + 2(32.0)(4) = 256 ft>s2
Ans.
Ans.
12–191. Solve
particle has an angular
$ Prob. 12–190 if the
#
acceleration u = 5 rad>s2 when u = 4 rad>s at u = p>2 rad.
y
Time Derivatives: Here,
p
r = 8u = 8 a b = 4p ft
2
$
$
r = 8 u = 8(5) = 40 ft>s2
r ⫽ (8 u) ft
#
#
r = 8 u = 8(4) = 32.0 ft>s
r
u
x
Velocity: Applying Eq. 12–25, we have
#
yr = r = 32.0 ft>s
#
yu = r u = 4p(4) = 50.3 ft>s
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
#
$
ar = r - r u2 = 40 - 4p A 42 B = -161 ft>s2
$
# #
au = r u + 2r u = 4p(5) + 2(32.0)(4) = 319 ft>s2
Ans.
Ans.
150
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*12–192. The boat moves along a path defined by
r2 = 310(103) cos 2u4 ft2, where u is in radians. If
u = (0.4t2) rad, where t is in seconds, determine the radial
and transverse components of the boat’s velocity and
acceleration at the instant t = 1 s.
$
#
Time Derivatives: Here, u = 0.4t , u = 0.8 t and u = 0.8 rad>s2. When t = 1 s,
#
u = 0.4 A 12 B = 0.4 rad and u = 0.8 (1) = 0.8 rad>s.
2
r2 = 10 A 103 B cos 2u
r = 1002cos 2u
#
#
2rr = -20 A 103 B sin 2uu
10(102) sin 2u #
#
r = u
r
#
$
$
#
2rr + 2r2 = -20 A 103 B A 2 cos 2uu2 + sin 2uu B
#
$
#
-10(103) A 2 cos 2uu2 + sin 2uu B - r 2
$
r =
r
10(10 ) sin 0.8
#
At u = 0.4 rad, r = 100 2cos 0.8 = 83.47 ft, r = (0.8) = -68.75 ft>s
83.47
3
-10(103) C 2 cos 0.8(0.82) + sin 0.8(0.8) D - (-68.75)2
$
and r =
= -232.23 ft>s2.
83.47
Velocity: Applying Eq. 12–25, we have
#
yr = r = -68.8 ft>s
#
yu = ru = 83.47(0.8) = 66.8 ft>s
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
#
$
ar = r - r u2 = -232.23 - 83.47 A 0.82 B = -286 ft>s2
$
##
au = ru + 2ru = 83.47(0.8) + 2(-68.75)(0.8) = -43.2 ft>s2
Ans.
Ans.
•12–193. A car travels along a road, which for a short distance
is defined by r = (200>u) ft, where u is in radians. If it
maintains a constant speed of v = 35 ft>s, determine the radial
and transverse components of its velocity when u = p>3 rad.
r =
200
600
2
=
ft
p
u u = p>3 rad
200 #
1800 #
#
r = - 2 u2
= - 2 u
u
p
u = p>3 rad
1800 #
#
yr = r = - 2 u
p
#
600 #
yu = ru =
u
p
y2 = y2r + y2u
352 = ¢ -
1800 # 2
600 # 2
u≤
u≤ + ¢
2
p
p
u = 0.1325 rad>s
yr = yu =
1800
(0.1325) = -24.2 ft>s
p2
Ans.
600
(0.1325) = 25.3 ft>s
p
Ans.
151
r
u
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12–194. For a short time the jet plane moves along a path
in the shape of a lemniscate, r2 = (2500 cos 2u) km2. At the
instant
u = 30°, the radar
$ tracking device is rotating at
#
u = 5(10 - 3) rad>s with u = 2(10 - 3) rad>s2. Determine the
radial and transverse components of velocity and
acceleration of the plane at this instant.
r2 ⫽ 2500 cos 2 u
r
u
#
$
Time Derivatives: Here, u = 5(10-3 B rad>s and u = 2(10-3) rad>s2.
r2 = 2500 cos 2u
r = 50 2cos 2u
#
#
2rr = -5000 sin 2uu
2500 sin 2u #
#
u
r = r
#
$
$
#
2rr + 2r2 = -5000 A 2 cos 2uu2 + sin 2uu B
#
$
#
-2500(2 cos 2 uu2 + sin 2 uu B - r 2
$
r =
r
2500 sin 60°
#
At u = 30°, r = 50 2cos 60° = 35.36 km, r = C 5 A 10-3 B D = -0.3062 km>s
35.36
-2500{2 cos 60° C 5(10-3) D 2 + sin 60° C 2(10-3) D } - ( -0.3062)2
$
= -0.1269 km>s2.
and r =
35.36
Velocity: Applying Eq. 12–25, we have
#
yr = r = -0.3062 km>s = 306 m>s
#
yu = ru = 35.36 C 5 A 10-3 B D = 0.1768 km>s = 177 m>s
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
#
$
ar = r - r u2 = -0.1269 - 35.36 C 5 A 10-3 B D 2
= -0.1278 km>s2 = -128 m>s2
$
##
a u = ru + 2ru = 35.36 C 2 A 10-3 B D + 2(-0.3062) C 5 A 10-3 B D
= 0.06765 km>s2 = 67.7 m>s2
Ans.
Ans.
152
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12–195. The mine car C is being pulled up the incline
using the motor M and the rope-and-pulley arrangement
shown. Determine the speed vP at which a point P on the
cable must be traveling toward the motor to move the car
up the plane with a constant speed of v = 2 m>s.
M
2sC + (sC - sP) = l
P
Thus,
v
3vC - vP = 0
C
Hence,
vP = 3( -2) = -6 m>s = 6 m>s Q
Ans.
*12–196. Determine the displacement of the log if the
truck at C pulls the cable 4 ft to the right.
C
B
2sB + (sB - sC) = l
3sB - sC = l
3¢sB - ¢sC = 0
Since ¢sC = -4, then
3¢sB = -4
¢sB = -1.33 ft = 1.33 ft :
Ans.
•12–197. If the hydraulic cylinder H draws in rod BC at
2 ft>s, determine the speed of slider A.
A
B
H
2sH + sA = l
2vH = -vA
2(2) = -vA
vA = -4 ft>s = 4 ft>s ;
Ans.
153
C
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12–198. If end A of the rope moves downward with a speed
of 5 m>s, determine the speed of cylinder B.
Position Coordinates: By referring to Fig. a, the length of the two ropes written in
terms of the position coordinates sA, sB, and sC are
sB + 2a + 2sC = l1
sB + 2sC = l1 - 2a
(1)
and
A
5 m/s
sA + (sA - sC) = l2
2sA - sC = l2
(2)
B
Eliminating sC from Eqs. (1) and (2) yields
sB + 4sA = l1 - 2a + 2l2
Time Derivative: Taking the time derivative of the above equation,
A+TB
vB + 4vA = 0
Here, vA = 5 m>s. Thus,
vB + 4(5) = 0
vB = -20 m>s = 20 m>s c
Ans.
B
12–199. Determine the speed of the elevator if each
motor draws in the cable with a constant speed of 5 m>s.
Position Coordinates: By referring to Fig. a, the length of the two cables written in
terms of the position coordinates are
sE + (sE - sA) + sC = l1
2sE - sA + sC = l1
(1)
and
(sE - sB) + 2(sE - sC) = l2
3sE - sB - 2sC = l2
(2)
Eliminating sC from Eqs. (1) and (2) yields
7sE - 2sA - sB = 2l1 + l2
Time Derivative: Taking the time derivative of the above equation,
A+TB
7vE - 2vA - vB = 0
Here, vA = vB = -5 m>s. Thus,
7vE - C 2 A -5 B D -
A -5 B = 0
vE = -2.14 m>s = 2.14 m>s c
Ans.
154
C
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*12–200. Determine the speed of cylinder A, if the rope is
drawn towards the motor M at a constant rate of 10 m>s.
Position Coordinates: By referring to Fig. a, the length of the rope written in terms
of the position coordinates sA and sM is
3sA + sM = l
Time Derivative: Taking the time derivative of the above equation,
A+TB
A
3vA + vM = 0
M
h
Here, vM = 10 m>s. Thus,
3vA + 10 = 0
vA = -3.33 m>s = 3.33 m>s c
Ans.
•12–201. If the rope is drawn towards the motor M at a
speed of vM = (5t3>2) m>s, where t is in seconds, determine
the speed of cylinder A when t = 1 s.
Position Coordinates: By referring to Fig. a, the length of the rope written in terms
of the position coordinates sA and sM is
3sA + sM = l
Time Derivative: Taking the time derivative of the above equation,
A+TB
A
3vA + vM = 0
h
Here, vM = A 5t3>2 B m>s. Thus,
3vA + 5t3>2 = 0
5
5
vA = ¢ - t3>2 ≤ m>s = ¢ t3>2 ≤ m>s 2
= 1.67 m>s c
3
3
t=1 s
155
Ans.
M
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12–202. If the end of the cable at A is pulled down with a
speed of 2 m>s, determine the speed at which block B rises.
D
C
Position-Coordinate Equation: Datum is established at fixed pulley D. The position
of point A, block B and pulley C with respect to datum are sA, sB and sC,
respectively. Since the system consists of two cords, two position-coordinate
equations can be developed.
2sC + sA = l1
[1]
sB + A sB - sC B = l2
[2]
A
2 m /s
Eliminating sC from Eqs. [1] and [2] yields
B
sA + 4sB = l1 + 2l2
Time Derivative: Taking the time derivative of the above equation yields
yA + 4yB = 0
[3]
Since yA = 2 m>s, from Eq. [3]
A+TB
2 + 4yB = 0
yB = -0.5 m>s = 0.5 m>s c
Ans.
12–203. Determine the speed of B if A is moving
downwards with a speed of vA = 4 m>s at the instant shown.
Position Coordinates: By referring to Fig. a, the length of the two ropes written in
terms of the position coordinates sA, sB, and sC are
sA + 2sC = l1
(1)
B
and
sB + (sB - sC) = l2
(2)
Eliminating sC from Eqs. (1) and (2),
4 sB + sA = 2l2 + l1
(3)
Time Derivative: Taking the time derivative of Eq. (3),
A+TB
4vB + vA = 0
Here, vA = 4 m>s. Thus,
4vB + 4 = 0
vB = -1 m>s = 1 m>s c
Ans.
156
A
vA ⫽ 4 m/s
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*12–204. The crane is used to hoist the load. If the motors
at A and B are drawing in the cable at a speed of 2 ft>s and
4 ft>s, respectively, determine the speed of the load.
2 ft/s
A
B
4 ft/s
Position-Coordinate Equation: Datum is established as shown. The positon of point
A and B and load C with respect to datum are sA, sB and sC, respectively.
4sC + sA + sB + 2h = l
Time Derivative: Since h is a constant, taking the time derivative of the above
equation yields
4yC + yA + yB = 0
[1]
Since yA = 2 ft>s and yB = 4 ft>s, from Eq. [1]
4yC + 2 + 4 = 0
yC = -1.50 ft>s = 1.50 ft>s c
Ans.
•12–205. The cable at B is pulled downwards at 4 ft>s, and
the speed is decreasing at 2 ft>s2. Determine the velocity
and acceleration of block A at this instant.
D
2sA + (h - sC) = l
2vA = vC
A
C
sC + (sC - sB) = l
2vC = vB
B
vB = 4vA
aB = 4aA
Thus,
-4 = 4vA
vA = -1 ft>s = 1 ft>s c
Ans.
2 = 4aA
aA = 0.5 ft>s = 0.5 ft>s2 T
Ans.
157
4 ft/s
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12–206. If block A is moving downward with a speed of 4 ft>s
while C is moving up at 2 ft>s, determine the speed of block B.
sA + 2sB + sC = l
vA + 2vB + vC = 0
B
4 + 2vB - 2 = 0
C
A
vB = -1 ft>s = 1 ft>s c
Ans.
12–207. If block A is moving downward at 6 ft>s while block
C is moving down at 18 ft>s, determine the speed of block B.
sA = 2sB + sC = l
vA = 2vB + vC = 0
B
6 + 2vB + 18 = 0
C
A
vB = -12 ft/s = 12 ft>s c
Ans.
*12–208. If the end of the cable at A is pulled down with a
speed of 2 m>s, determine the speed at which block E rises.
Position-Coordinate Equation: Datum is established at fixed pulley. The position of
point A, pulley B and C and block E with respect to datum are sA, sB, sC and sE,
respectively. Since the system consists of three cords, three position-coordinate
equations can be developed.
2sB + sA = l1
[1]
sC + (sC - sB) = l2
[2]
sE + (sE - sC) = l3
[3]
2 m/s
A
C
D
E
Eliminating sC and sB from Eqs. [1], [2] and [3], we have
sA + 8sE = l1 + 2l2 + 4l3
Time Derivative: Taking the time derivative of the above equation yields
yA + 8yE = 0
[4]
Since yA = 2 m>s, from Eq. [3]
A+TB
B
2 + 8yE = 0
yE = -0.250 m>s = 0.250 m>s c
Ans.
158
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•12–209. If motors at A and B draw in their attached
cables with an acceleration of a = (0.2t) m>s2, where t is in
seconds, determine the speed of the block when it reaches a
height of h = 4 m, starting from rest at h = 0. Also, how
much time does it take to reach this height?
D
sA + 2sD = l
C
(sC - sD) + sC + sB = l¿
¢sA = -2¢sD
2¢sC - ¢sD + ¢sB = 0
If ¢sC = -4 and ¢sA = ¢sB, then,
¢sA = -2¢sD
2(-4) - ¢sD + ¢sA = 0
¢sD = -2.67 m
¢sA = ¢sB = 5.33 m
A
Thus,
vA = -2vD
(1)
2vC - vD + vB = 0
(2)
a = 0.2t
dv = a dt
L0
v
dv =
L0
t
0.2t dt
v = 0.1t2
ds = v dt
L0
s
s =
ds =
L0
t
0.1t2 dt
0.1 3
t = 5.33
3
t = 5.428 s = 5.43 s
Ans.
v = 0.1(5.428)2 = 2.947 m>s
vA = vB = 2.947 m>s
Thus, from Eqs. (1) and (2):
2.947 = -2vD
vD = -1.474
2vC - (-1.474) + 2.947 = 0
vC = -2.21 m>s = 2.21 m>s c
Ans.
159
h
B
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12–210. The motor at C pulls in the cable with an
acceleration aC = (3t2) m>s2, where t is in seconds. The
motor at D draws in its cable at a D = 5 m>s2. If both motors
start at the same instant from rest when d = 3 m, determine
(a) the time needed for d = 0, and (b) the velocities of
blocks A and B when this occurs.
D
B
A
d⫽3m
For A:
sA + (sA - sC) = l
2vA = vC
2aA = aC = -3t2
aA = -1.5t2 = 1.5t2 :
vA = 0.5t3 :
sA = 0.125t4 :
For B:
aB = 5 m>s2 ;
vB = 5t ;
sB = 2.5t2 ;
Require sA + sB = d
0.125t4 + 2.5t2 = 3
Set u = t2
0.125u2 + 2.5u = 3
The positive root is u = 1.1355. Thus,
t = 1.0656 = 1.07 s
Ans.
vA = .0.5(1.0656)2 = 0.605 m>s
Ans.
vB = 5(1.0656) = 5.33 m>s
Ans.
160
C
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12–211. The motion of the collar at A is controlled by a
motor at B such that when the collar is at sA = 3 ft it is
moving upwards at 2 ft>s and decreasing at 1 ft>s2.
Determine the velocity and acceleration of a point on the
cable as it is drawn into the motor B at this instant.
2s2A
4 ft
sA
2
+ 4 + sB = l
1
1 2
#
A sA + 16 B - 2 (2sA)sA + sB = 0
2
A
1
#
#
sB = -sAsA A s2A + 16 B - 2
B
1
1
3
1
# 2
#
$
#
#
sB = - B asA b A sA2 + 16 B - 2 + sAsA A sA2 + 16 B - 2 + sAsA a- b A s2A + 16 B - 2 a2sAsA b R
2
$
#
A sAsA B 2
A sA B 2 + sAsA
#
sB =
3 1
A s2A + 16 B 2
A s2A + 16 B 2
Evaluating these equations:
sB = -3(-2) A (3)2 + 16 B - 2 = 1.20 ft>s T
1
sB =
((3)( -2))2
3
2
((3)2 + 16)
-
(-2)2 + 3(1)
1
((3)2 + 16)2
Ans.
= -1.11 ft>s2 = 1.11 ft>s2 c
Ans.
*12–212. The man pulls the boy up to the tree limb C by
walking backward at a constant speed of 1.5 m>s.
Determine the speed at which the boy is being lifted at the
instant xA = 4 m. Neglect the size of the limb. When
xA = 0, yB = 8 m, so that A and B are coincident, i.e., the
rope is 16 m long.
C
yB
8m
B
we have lAC = 2x2A + 82. Thus,
Position-Coordinate Equation: Using the Pythagorean theorem to determine lAC,
A
l = lAC + yB
16 = 2x2A + 82 + yB
yB = 16 -
2x2A
xA
+ 64
[1]
dxA
Time Derivative: Taking the time derivative of Eq. [1] and realizing that yA =
dt
dyB
and yB =
, we have
dt
dyB
xA
dxA
= 2
dt
2xA + 64 dt
xA
yA
yB = 2
2xA + 64
yB =
[2]
At the instant xA = 4 m, from Eq. [2]
yB = -
4
24 + 64
2
(1.5) = -0.671 m>s = 0.671 m>s c
Ans.
Note: The negative sign indicates that velocity yB is in the opposite direction to that
of positive yB.
161
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•12–213. The man pulls the boy up to the tree limb C by
walking backward. If he starts from rest when xA = 0 and
moves backward with a constant acceleration aA = 0.2 m>s2,
determine the speed of the boy at the instant yB = 4 m.
Neglect the size of the limb.When xA = 0, yB = 8 m, so that A
and B are coincident, i.e., the rope is 16 m long.
C
yB
8m
B
A
Position-Coordinate Equation: Using the Pythagorean theorem to determine lAC,
we have lAC = 2x2A + 82. Thus,
xA
l = lAC + yB
16 = 2x2A + 82 + yB
yB = 16 - 2x2A + 64
[1]
dxA
Time Derivative: Taking the time derivative of Eq. [1] Where yA =
and
dt
dyB
yB =
, we have
dt
dyB
xA
dxA
= 2
dt
2xA + 64 dt
xA
yB = yA
2
2xA + 64
yB =
[2]
At the instant yB = 4 m, from Eq. [1], 4 = 16 - 2x2A + 64, xA = 8.944 m. The
velocity of the man at that instant can be obtained.
y2A = (y0)2A + 2(ac)A C sA - (s0)A D
y2A = 0 + 2(0.2)(8.944 - 0)
yA = 1.891 m>s
Substitute the above results into Eq. [2] yields
yB = -
8.944
28.9442 + 64
(1.891) = -1.41 m>s = 1.41 m>s c
Ans.
Note: The negative sign indicates that velocity yB is in the opposite direction to that
of positive yB.
162
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12–214. If the truck travels at a constant speed of
vT = 6 ft>s, determine the speed of the crate for any angle u
of the rope. The rope has a length of 100 ft and passes over
a pulley of negligible size at A. Hint: Relate the coordinates
xT and xC to the length of the rope and take the time
derivative. Then substitute the trigonometric relation
between xC and u.
xC
xT
T
A
20 ft
C
u
2(20)2 + x2C + xT = l = 100
1
1
#
#
A (20)2 + (xC)2 B - 2 a2xCxC b + xT = 0
2
#
#
Since xT = vT = 6 ft>s, vC = xC, and
xC = 20 ctn u
Then,
(20 ctn u)vC
1
(400 + 400 ctn2 u)2
= -6
Since 1 + ctn2 u = csc2 u,
a
ctn u
b v = cos uvC = -6
csc u C
vC = -6 sec u = (6 sec u) ft>s :
Ans.
12–215. At the instant shown, car A travels along the
straight portion of the road with a speed of 25 m>s. At this
same instant car B travels along the circular portion of the
road with a speed of 15 m>s. Determine the velocity of car B
relative to car A.
15⬚
r ⫽ 200 m
Velocity: Referring to Fig. a, the velocity of cars A and B expressed in Cartesian
vector form are
B
vB = [15 cos 15° i - 15 sin 15° j] m>s = [14.49i - 3.882j] m>s
Applying the relative velocity equation,
vB = vA + vB>A
14.49i - 3.882j = 21.65i - 12.5j + vB>A
vB>A = [-7.162i + 8.618j] m>s
Thus, the magnitude of vB/A is given by
Ans.
The direction angle uv of vB/A measured from the x axis, Fig. a is
uv = tan - 1 a
8.618
b = 50.3°
7.162
Ans.
163
C
30⬚
vA = [25 cos 30° i - 25 sin 30° j] m>s = [21.65i - 12.5j] m>s
vB>A = 2(-7.162)2 + 8.6182 = 11.2 m>s
A
15⬚
vT
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*12–216. Car A travels along a straight road at a speed of
25 m>s while accelerating at 1.5 m>s2. At this same instant
car C is traveling along the straight road with a speed of
30 m>s while decelerating at 3 m>s2. Determine the velocity
and acceleration of car A relative to car C.
1.5 m/s
r ⫽ 100 m
15 m/s
Velocity: The velocity of cars A and B expressed in Cartesian vector form are
vA = [-25 cos 45°i - 25 sin 45°j] m>s = [-17.68i - 17.68j] m>s
vC = [-30j] m>s
Applying the relative velocity equation, we have
vA = vC + vA>C
-17.68i - 17.68j = -30j + vA>C
vA>C = [-17.68i + 12.32j] m>s
Thus, the magnitude of vA/C is given by
vA>C = 2(-17.68)2 + 12.322 = 21.5 m>s
Ans.
and the direction angle uv that vA/C makes with the x axis is
12.32
b = 34.9° b
17.68
Ans.
Acceleration: The acceleration of cars A and B expressed in Cartesian vector form are
aA = [-1.5 cos 45°i - 1.5 sin 45°j] m>s2 = [-1.061i - 1.061j] m>s2
aC = [3j]m>s2
Applying the relative acceleration equation,
aA = aC + aA>C
-1.061i - 1.061j = 3j + aA>C
aA>C = [-1.061i - 4.061j] m>s2
Thus, the magnitude of aA/C is given by
aA>C = 2(-1.061)2 + (-4.061)2 = 4.20 m>s2
Ans.
and the direction angle ua that aA/C makes with the x axis is
ua = tan - 1 a
45⬚
2
2 m/s2
uv = tan - 1 a
A
25 m/s
4.061
b = 75.4° d
1.061
Ans.
164
3 m/s2
30⬚
B
C
30 m/s
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•12–217. Car B is traveling along the curved road with a
speed of 15 m>s while decreasing its speed at 2 m>s2. At this
same instant car C is traveling along the straight road with a
speed of 30 m>s while decelerating at 3 m>s2. Determine the
velocity and acceleration of car B relative to car C.
1.5 m/s
r ⫽ 100 m
15 m/s
Velocity: The velocity of cars B and C expressed in Cartesian vector form are
vB = [15 cos 60° i - 15 sin 60° j] m>s = [7.5i - 12.99j] m>s
vC = [-30j] m>s
Applying the relative velocity equation,
vB = vC + vB>C
7.5i - 12.99j = -30j + vB>C
vB>C = [7.5i + 17.01j] m>s
Thus, the magnitude of vB/C is given by
vBC = 27.52 + 17.012 = 18.6 m>s
Ans.
and the direction angle uv that vB/C makes with the x axis is
17.01
b = 66.2°
7.5
Ans.
Acceleration: The normal component of car B’s acceleration is (aB)n =
vB 2
r
152
= 2.25 m>s2. Thus, the tangential and normal components of car B’s
100
acceleration and the acceleration of car C expressed in Cartesian vector form are
=
(aB)t = [-2 cos 60° i + 2 sin 60°j] = [-1i + 1.732j] m>s2
(aB)n = [2.25 cos 30° i + 2.25 sin 30° j] = [1.9486i + 1.125j] m>s2
aC = [3j] m>s2
Applying the relative acceleration equation,
aB = aC + aB>C
(-1i + 1.732j) + (1.9486i + 1.125j) = 3j + aB>C
aB>C = [0.9486i - 0.1429j] m>s2
Thus, the magnitude of aB/C is given by
aB>C = 20.94862 + (-0.1429)2 = 0.959 m>s2
Ans.
and the direction angle ua that aB/C makes with the x axis is
ua = tan - 1 a
45⬚
2
2 m/s2
uv = tan - 1 a
A
25 m/s
0.1429
b = 8.57°
0.9486
Ans.
165
3 m/s2
30⬚
B
C
30 m/s
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12–218. The ship travels at a constant speed of vs = 20 m>s
and the wind is blowing at a speed of vw = 10 m>s, as shown.
Determine the magnitude and direction of the horizontal
component of velocity of the smoke coming from the smoke
stack as it appears to a passenger on the ship.
vs ⫽ 20 m/s
30⬚
vw ⫽ 10 m/s
45⬚
y
x
Solution I
Vector Analysis: The velocity of the smoke as observed from the ship is equal to the
velocity of the wind relative to the ship. Here, the velocity of the ship and wind
expressed in Cartesian vector form are vs = [20 cos 45° i + 20 sin 45° j] m>s
= [14.14i + 14.14j] m>s and vw = [10 cos 30° i - 10 sin 30° j] = [8.660i - 5j] m>s.
Applying the relative velocity equation,
vw = vs + vw>s
8.660i - 5j = 14.14i + 14.14j + vw>s
vw>s = [-5.482i - 19.14j] m>s
Thus, the magnitude of vw/s is given by
vw = 2( -5.482)2 + (-19.14)2 = 19.9m>s
Ans.
and the direction angle u that vw/s makes with the x axis is
u = tan - 1 a
19.14
b = 74.0°
5.482
Ans.
Solution II
Scalar Analysis: Applying the law of cosines by referring to the velocity diagram
shown in Fig. a,
vw>s = 2202 + 102 - 2(20)(10) cos 75°
= 19.91 m>s = 19.9 m>s
Ans.
Using the result of vw/s and applying the law of sines,
sin f
sin 75°
=
10
19.91
f = 29.02°
Thus,
u = 45° + f = 74.0°
Ans.
166
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12–219. The car is traveling at a constant speed of 100 km>h.
If the rain is falling at 6 m>s in the direction shown, determine
the velocity of the rain as seen by the driver.
30⬚
Solution I
1h
km 1000 m
ba
ba
b
h
1 km
3600 s
= 27.78 m>s.The velocity of the car and the rain expressed in Cartesian vector form
are vc = [-27.78i] m>s and vr = [6 sin 30°i - 6 cos 30°j] = [3i - 5.196j] m>s.
Applying the relative velocity equation, we have
Vector Analysis: The speed of the car is vc = a100
vr = vc + vr>c
3i - 5.196j = -27.78i + vr>c
vr>c = [30.78i - 5.196j] m>s
Thus, the magnitude of vr/c is given by
vr>c = 230.782 + ( -5.196)2 = 31.2m>s
Ans.
and the angle xr/c makes with the x axis is
u = tan - 1 a
5.196
b = 9.58°
30.78
Ans.
Solution II
Scalar Analysis: Referring to the velocity diagram shown in Fig. a and applying the
law of cosines,
vr>c = 227.782 + 62 - 2(27.78)(6) cos 120°
= 19.91 m>s = 19.9 m>s
Ans.
Using the result of vr/c and applying the law of sines,
sin u
sin 120°
=
6
31.21
u = 9.58°
Ans.
167
vr
vC ⫽ 100 km/h
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*12–220. The man can row the boat in still water with a
speed of 5 m>s. If the river is flowing at 2 m>s, determine
the speed of the boat and the angle u he must direct the
boat so that it travels from A to B.
B
vw ⫽ 2 m/s
5 m/s
u
Solution I
A
Vector Analysis: Here, the velocity vb of the boat is directed from A to B. Thus,
f = tan - 1 a
50
b = 63.43°. The magnitude of the boat’s velocity relative to the
25
flowing river is vb>w = 5 m>s. Expressing vb, vw, and vb/w in Cartesian vector form,
we have vb = vb cos 63.43i + vb sin 63.43j = 0.4472vb i + 0.8944vb j, vw = [2i] m>s,
and vb>w = 5 cos ui + 5 sin uj. Applying the relative velocity equation, we have
vb = vw + vb>w
0.4472vb i + 0.8944vb j = 2i + 5 cos ui + 5 sin uj
0.4472vb i + 0.8944vb j = (2 + 5 cos u)i + 5 sin uj
Equating the i and j components, we have
0.4472vb = 2 + 5 cos u
(1)
0.8944vb = 5 sin u
(2)
Solving Eqs. (1) and (2) yields
vb = 5.56 m>s
u = 84.4°
Ans.
Solution II
Scalar Analysis: Referring to the velocity diagram shown in Fig. a and applying the
law of cosines,
52 = 22 + vb 2 - 2(2)(vb) cos 63.43°
vb 2 - 1.789vb - 21 = 0
vb =
-(-1.789) ; 2(-1.789)2 - 4(1)(-21)
2(1)
Choosing the positive root,
nb = 5.563 m>s = 5.56 m>s
Ans.
Using the result of nb and applying the law of sines,
sin 63.43°
sin 180° - u
=
5.563
5
u = 84.4°
Ans.
168
25 m
50 m
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•12–221. At the instant shown, cars A and B travel at speeds
of 30 mi>h and 20 mi>h, respectively. If B is increasing its
speed by 1200 mi>h2, while A maintains a constant speed,
determine the velocity and acceleration of B with respect to A.
B
30⬚
0.3 mi
vB ⫽ 20 mi/h
vA ⫽ 30 mi/h
vB = vA + vB>A
20 30°
h
= 30 + (vB/A)x + (vB/A )y
;
+ B
A:
A+cB
c
:
-20 sin 30° = -30 + (vB>A)x
20 cos 30° = (vB>A)y
Solving
(vB>A)x = 20 :
(vB>A)y = 17.32 c
vB>A = 2(20)2 + (17.32)2 = 26.5 mi>h
u = tan - 1(
(aB)n =
Ans.
17.32
) = 40.9° au
20
Ans.
(20)2
= 1333.3
0.3
aB = aA + aB>A
1200 30°
h
1333.3
+ au = 0 + (aB>A )x + (aB>A)y
30°
+ B
A:
A+cB
:
c
-1200 sin 30° + 1333.3 cos 30° = (aB>A)x
1200 cos 30° + 1333.3 sin 30° = (aB>A)y
Solving
(aB>A)x = 554.7 :
;
(aB>A)y = 1705.9 c
aB>A = 2(554.7)2 + 1705.9)2 = 1.79(103) mi>h2
u = tan - 1(
Ans.
1705.9
) = 72.0° au
554.7
Ans.
169
A
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12–222. At the instant shown, cars A and B travel at
speeds of 30 m>h and 20 mi>h, respectively. If A is
increasing its speed at 400 mi>h2 whereas the speed of B is
decreasing at 800 mi>h2, determine the velocity and
acceleration of B with respect to A.
B
0.3 mi
vB ⫽ 20 mi/h
vA ⫽ 30 mi/h
vB = vA + vB>A
20 30°
h
= 30 + (vB/A)x + (vB/A )y
+ B
A:
A+cB
c
:
;
-20 sin 30° = -30 + (vB>A)x
20 cos 30° = (vB>A)y
Solving
(vB>A)x = 20 :
(vB>A)y = 17.32 c
vB>A = 2(20)2 + (17.32)2 = 26.5 mi>h
u = tan - 1(
Ans.
17.32
) = 40.9° au
20
Ans.
aB + aA + aB>A
[
202
= 1333.3]
0.3
au
A
+
:
B
+ [800]
30°
30° f
= [400] + [(aB>A)x] + [(aB>A)y]
;
:
c
1333.3 cos 30° + 800 sin 30° = -400 + (aB>A)x
(aB>A)x = 1954.7 :
A+cB
1333.3 sin 30° - 800 cos 30° = (aB>A)y
(aB>A)y = -26.154 = 26.154 T
(aB>A) = 2(1954.7)2 + (26.154)2
aB>A = 1955 mi>h2
u = tan - 1(
30⬚
Ans.
26.154
) = 0.767° c u
1954.7
Ans.
170
A
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12–223. Two boats leave the shore at the same time and travel
in the directions shown. If vA = 20 ft>s and vB = 15 ft>s,
determine the velocity of boat A with respect to boat B. How
long after leaving the shore will the boats be 800 ft apart?
vA
A
B
vB
vA = vB + vA>B
-20 sin 30°i + 20 cos 30°j = 15 cos 45°i + 15 sin 45°j + vA>B
30⬚
vA>B = {-20.61i + 6.714j} ft>s
O
vA>B = 2(-20.61)2 + (+6.714)2 = 21.7 ft>s
u = tan - 1 (
45⬚
Ans.
6.714
) = 18.0° b
20.61
Ans.
(800)2 = (20 t)2 + (15 t)2 - 2(20 t)(15 t) cos 75°
t = 36.9 s
Ans.
Also
t =
800
800
=
= 36.9 s
vA>B
21.68
Ans.
*12–224. At the instant shown, cars A and B travel at speeds
of 70 mi>h and 50 mi>h, respectively. If B is increasing its speed
by 1100 mi>h2, while A maintains a constant speed, determine
the velocity and acceleration of B with respect to A. Car B
moves along a curve having a radius of curvature of 0.7 mi.
A
vA ⫽ 70 mi/h
Relative Velocity:
vB = vA + vB>A
30⬚
50 sin 30°i + 50 cos 30°j = 70j + vB>A
vB>A = {25.0i - 26.70j} mi>h
Thus, the magnitude of the relative velocity vB/A is
yB>A = 225.02 + (-26.70)2 = 36.6 mi>h
Ans.
The direction of the relative velocity is the same as the direction of that for relative
acceleration. Thus
u = tan - 1
26.70
= 46.9° c
25.0
Ans.
Relative Acceleration: Since car B is traveling along a curve, its normal
y2B
502
= 3571.43 mi>h2. Applying Eq. 12–35 gives
=
acceleration is (aB)n =
r
0.7
aB = aA + aB>A
(1100 sin 30° + 3571.43 cos 30°)i + (1100 cos 30° - 3571.43 sin 30°)j = 0 + aB>A
aB>A = {3642.95i - 833.09j} mi>h2
Thus, the magnitude of the relative velocity aB/A is
aB>A = 23642.952 + (-833.09)2 = 3737 mi>h2
Ans.
And its direction is
f = tan - 1
vB ⫽ 50 mi/h
833.09
= 12.9° c
3642.95
Ans.
171
B
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•12–225. At the instant shown, cars A and B travel at
speeds of 70 mi>h and 50 mi>h, respectively. If B is
decreasing its speed at 1400 mi>h2 while A is increasing its
speed at 800 mi>h2, determine the acceleration of B with
respect to A. Car B moves along a curve having a radius of
curvature of 0.7 mi.
A
vA ⫽ 70 mi/h
Relative Acceleration: Since car B is traveling along a curve, its normal acceleration
y2B
502
is (aB)n =
=
= 3571.43 mi>h2. Applying Eq. 12–35 gives
r
0.7
vB ⫽ 50 mi/h
B
30⬚
aB = aA + aB>A
(3571.43 cos 30° - 1400 sin 30°)i + (-1400 cos 30° - 3571.43 sin 30°)j = 800j + aB>A
aB>A = {2392.95i - 3798.15j} mi>h2
Thus, the magnitude of the relative acc. aB/A is
aB>A = 22392.952 + (-3798.15)2 = 4489 mi>h2
Ans.
And its direction is
f = tan - 1
3798.15
= 57.8° c
2392.95
Ans.
12–226. An aircraft carrier is traveling forward with a
velocity of 50 km>h. At the instant shown, the plane at A
has just taken off and has attained a forward horizontal air
speed of 200 km>h, measured from still water. If the plane
at B is traveling along the runway of the carrier at
175 km>h in the direction shown, determine the velocity of
A with respect to B.
B
15⬚
vB = vC + vB>C
50 km/h
v B = 50i + 175 cos 15°i + 175 sin 15°j = 219.04i + 45.293j
vA = vB + vA>B
200i = 219.04i + 45.293j + (vA>B)xi + (vA>B)y j
200 = 219.04 + (vA>B)x
0 = 45.293 + (vA>B)y
(vA>B)x = -19.04
(vA>B)y = -45.293
vA>B = 2(-19.04)2 + (-45.293)2 = 49.1 km>h
u = tan - 1 a
A
Ans.
45.293
b = 67.2° d
19.04
Ans.
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12–227. A car is traveling north along a straight road at
50 km>h. An instrument in the car indicates that the wind is
directed towards the east. If the car’s speed is 80 km>h, the
instrument indicates that the wind is directed towards the
north-east. Determine the speed and direction of the wind.
Solution I
Vector Analysis: For the first case, the velocity of the car and the velocity of the wind
relative to the car expressed in Cartesian vector form are vc = [50j] km>h and
vW>C = (vW>C)1 i. Applying the relative velocity equation, we have
vc = vw + vw>c
vw = 50j + (vw>c)1 i
vw = (vw>c)1i + 50j
(1)
For the second case, vC = [80j] km>h and vW>C = (vW>C)2 cos 45°i + (vW>C)2 sin 45° j.
Applying the relative velocity equation, we have
vw = vc + vw>c
vw = 80j + (vw>c)2 cos 45°i + (vw>c)2 sin 45° j
vw = (vw>c)2 cos 45° i + C 80 + (vw>c)2 sin 45° D j
(2)
Equating Eqs. (1) and (2) and then the i and j components,
(vw>c)1 = (vw>c)2 cos 45°
(3)
50 = 80 + (vw>c)2 sin 45°
(4)
Solving Eqs. (3) and (4) yields
(vw>c)2 = -42.43 km>h
(vw>c)1 = -30 km>h
Substituting the result of (vw>c)1 into Eq. (1),
vw = [-30i + 50j] km>h
Thus, the magnitude of vW is
vw = 2(-30)2 + 502 = 58.3 km>h
Ans.
and the directional angle u that vW makes with the x axis is
u = tan - 1 a
50
b = 59.0° b
30
Ans.
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*12–228. At the instant shown car A is traveling with a
velocity of 30 m>s and has an acceleration of 2 m>s2 along
the highway. At the same instant B is traveling on the
trumpet interchange curve with a speed of 15 m>s, which is
decreasing at 0.8 m>s2. Determine the relative velocity and
relative acceleration of B with respect to A at this instant.
A
B
vB = vA + vB>A
60⬚
15 cos 60°i + 15 sin 60°j = 30i + (vB>A)xi + (vB>A)y j
15 cos 60° = 30 + (vB>A)x
15 sin 60° = 0 + (vB>A)y
(vB>A)x = -22.5 = 22.5 m>s ;
(vB>A)y = 12.99 m>s c
vB>A = 2(22.5)2 + (12.99)2 = 26.0 m>s
u = tan - 1 a
(aB)n =
Ans.
12.99
b = 30° b
22.5
Ans.
v2
152
= 0.9 m>s2
=
r
250
aB = aA + aB>A
-0.8 cos 60°i - 0.8 sin 60°j + 0.9 sin 60°i - 0.9 cos 60°j = 2i + (aB>A)x i + (aB>A)y j
-0.8 cos 60° + 0.9 sin 60° = 2 + (aB>A)x
-0.8 sin 60° - 0.9 cos 60° = (aB>A)y
(aB>A)x = -1.6206 ft>s2 = 1.6206 m>s2 ;
(aB>A)y = -1.1428 ft>s2 = 1.1428 m>s2 T
aB>A = 2(1.6206)2 + (1.1428)2 = 1.98 m>s2
f = tan - 1 ¢
Ans.
1.1428
≤ = 35.2° d
1.6206
Ans.
174
r ⫽ 250 m
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•12–229. Two cyclists A and B travel at the same constant
speed v. Determine the velocity of A with respect to B if A
travels along the circular track, while B travels along the
diameter of the circle.
v
A
r
vA = y sin ui + y cos u j
f
vB = y i
B
u
v
vA>B = vA - vB
= (y sin ui + y cos u j) - y i
= (y sin u - y)i + y cos uj
yA>B = 2(y sin u - y)2 + (y cos u)2
= 22y2 - 2y2 sin u
= y22(1 - sin u)
Ans.
12–230. A man walks at 5 km>h in the direction of a
20-km>h wind. If raindrops fall vertically at 7 km>h in still air,
determine the direction in which the drops appear to fall with
respect to the man. Assume the horizontal speed of the
raindrops is equal to that of the wind.
vw ⫽ 20 km/h
Relative Velocity: The velocity of the rain must be determined first. Applying
Eq. 12–34 gives
vm ⫽ 5 km/h
vr = vw + vr>w = 20i + (-7j) = {20i - 7j} km>h
Thus, the relative velocity of the rain with respect to the man is
vr = vm + vr>m
20i - 7j = 5i + vr>m
vr>m = {15i - 7j} km>h
The magnitude of the relative velocity vr/m is given by
yr>m = 2152 + (-7)2 = 16.6 km>h
Ans.
And its direction is given by
u = tan - 1
7
= 25.0° c
15
Ans.
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12–231. A man can row a boat at 5 m>s in still water. He
wishes to cross a 50-m-wide river to point B, 50 m
downstream. If the river flows with a velocity of 2 m>s,
determine the speed of the boat and the time needed to
make the crossing.
50 m
2 m/s
A
u
50 m
Relative Velocity:
vb = vr + vb>r
B
yb sin 45°i - yb cos 45°j = -2j + 5 cos ui - 5 sin uj
Equating i and j component, we have
yb sin 45° = 5 cos u
[1]
-yb cos 45° = -2 - 5 sin u
[2]
Solving Eqs. [1] and [2] yields
u = 28.57°
yb = 6.210 m>s = 6.21 m>s
Ans.
Thus, the time t required by the boat to travel from point A to B is
t =
sAB
2502 + 502
= 11.4 s
=
yb
6.210
Ans.
176