12–1.
Starting from rest, a particle moving in a straight line has an acceleration of a = (bt + c). What
is the particle’s velocity at t1 and what is its position at t2?
Given:
b
2
m
3
c
6
s
m
s
2
t1
6s
t2
Solution:
t
a ( t)
bt c
v ( t)
´
µ a ( t ) dt
¶0
v t1
0
m
s
d t2
80.7 m
t
d ( t)
Ans.
´
µ v ( t ) dt
¶0
11 s
12–2.
When a train is traveling along a straight track at 2 m/s, it
begins to accelerate at a = 160 v-42 m>s2, where v is in m/s.
Determine its velocity v and the position 3 s after the
acceleration.
s
SOLUTION
a =
dv
dt
dt =
dv
a
v
3
dt =
L0
3 =
dv
-4
L2 60v
1
(v5 - 32)
300
v = 3.925 m>s = 3.93 m>s
Ans.
ads = vdv
ds =
s
L0
ds =
s =
1 5
vdv
=
v dv
a
60
1
60 L2
3.925
v5 dv
1 v6 3.925
a b`
60 6 2
= 9.98 m
Ans.
v
12–3.
The acceleration of a particle as it moves along a straight line is given by a = bt + c. If s = s0 and
v = v0 when t = 0, determine the particle's velocity and position when t = t1. Also, determine the
total distance the particle travels during this time period.
Given:
b
2
m
c
3
m
1
2
s
s0
1m
v
v0 v0
2
s
m
s
t1
6s
Solution:
v
´
µ 1 dv
¶v
0
s
´
µ 1 ds
¶s
0
t
2
´
µ ( b t c) dt
¶0
´
µ
µ
¶
t
0
2
·
§
bt
¨v0 c t¸ dt
2
©
¹
bt
ct
2
s
b 3 c 2
t t
6
2
s0 v0 t 2
When t = t1
b t1
v1
v0 s1
s0 v0 t1 2
c t1
b 3 c 2
t1 t1
6
2
v1
32.00
m
s
Ans.
s1
67.00 m
Ans.
The total distance traveled depends on whether the particle turned around or not. To tell we
will plot the velocity and see if it is zero at any point in the interval
2
t
0 0.01t1 t1
v0 v ( t)
bt
ct
2
d
40
v( t) 20
0
If v never goes to zero then
0
2
4
t
6
s1 s0
d
66.00 m
Ans.
12–4.
Traveling with an initial speed of 70 km>h, a car accelerates
at 6000 km>h2 along a straight road. How long will it take to
reach a speed of 120 km>h? Also, through what distance
does the car travel during this time?
SOLUTION
v = v1 + ac t
120 = 70 + 6000(t)
t = 8.33(10 - 3) hr = 30 s
Ans.
v2 = v21 + 2 ac(s - s1)
(120)2 = 702 + 2(6000)(s - 0)
s = 0.792 km = 792 m
Ans.
12–5.
A bus starts from rest with a constant acceleration of 1 m>s2.
Determine the time required for it to attain a speed of 25 m>s
and the distance traveled.
SOLUTION
Kinematics:
v0 = 0, v = 25 m>s, s0 = 0, and ac = 1 m>s2.
+ B
A:
v = v0 + act
25 = 0 + (1)t
t = 25 s
+ B
A:
Ans.
v2 = v02 + 2ac(s - s0)
252 = 0 + 2(1)(s - 0)
s = 312.5 m
Ans.
12–6.
A stone A is dropped from rest down a well, and in 1 s
another stone B is dropped from rest. Determine the
distance between the stones another second later.
SOLUTION
+ T s = s1 + v1 t +
sA = 0 + 0 +
1 2
a t
2 c
1
(9.81)(2)2
2
sA = 19.62 m
sA = 0 + 0 +
1
(9.81)(1)2
2
sB = 4.91 m
¢s = 19.62 - 4.91 = 14.71 m
Ans.
12–7.
A bicyclist starts from rest and after traveling along a
straight path a distance of 20 m reaches a speed of 30 km/h.
Determine his acceleration if it is constant. Also, how long
does it take to reach the speed of 30 km/h?
SOLUTION
v2 = 30 km>h = 8.33 m>s
v22 = v21 + 2 ac (s2 - s1)
(8.33)2 = 0 + 2 ac (20 - 0)
ac = 1.74 m>s2
Ans.
v2 = v1 + ac t
8.33 = 0 + 1.74(t)
t = 4.80 s
Ans.
12–8.
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.
SOLUTION
a =
5
A 3s3 + s2 B
1
5
a ds = v dv
2
v
5 ds
L1 A 3s + s
1
3
0.8351 =
5
2
B
=
L0
v dv
1 2
v
2
v = 1.29 m>s
Ans.
12–9.
If it takes 3 s for a ball to strike the ground when it is released
from rest, determine the height in meters of the building
from which it was released. Also, what is the velocity of the
ball when it strikes the ground?
SOLUTION
Kinematics:
v0 = 0, ac = g = 9.81 m>s2, t = 3 s, and s = h.
A+TB
v = v0 + act
= 0 + (9.81)(3)
= 29.4 m>s
A+TB
Ans.
1
s = s0 + v0t + act2
2
1
h = 0 + 0 + (9.81)(32)
2
= 44.1 m
Ans.
12–10.
A particle, initially at the origin, moves along a straight line through a fluid medium such that its
velocity is defined as v = b(1 - e-ct). Determine the displacement of the particle during the time
0 < t < t1.
Given:
b
1.8
m
s
c
0.3
s
t1
3s
Solution:
v ( t)
ct
b1e
t
sp ( t)
´
µ v ( t ) dt
¶0
sp t1
1.839 m
Ans.
12–1 1.
A truck traveling along a straight road at speed v1, increases its speed to v2 in time t. If its
acceleration is constant, determine the distance traveled.
Given:
v1
km
hr
20
v2
120
km
hr
t
15 s
Solution:
a
d
v2 v1
t
v1 t a
1.852
m
2
s
1 2
at
2
d
291.7 m
Ans.
12–12.
Determine the time required for a car to travel 1 km along
a road if the car starts from rest, reaches a maximum speed
at some intermediate point, and then stops at the end of
the road. The car can accelerate at 1.5 m>s2 and decelerate
at 2 m>s2.
SOLUTION
Using formulas of constant acceleration:
v2 = 1.5 t1
x =
1
(1.5)(t21)
2
0 = v2 - 2 t2
1
(2)(t22)
2
1000 - x = v2t2 -
Combining equations:
t1 = 1.33 t2; v2 = 2 t2
x = 1.33 t22
1000 - 1.33 t22 = 2 t22 - t22
t2 = 20.702 s;
t1 = 27.603 s
t = t1 + t2 = 48.3 s
Ans.
12–13.
The acceleration of a rocket traveling upward is given by ap = b + c sp. Determine the rocket’s
velocity when sp = sp1 and the time needed to reach this altitude. Initially, vp = 0 and sp = 0 when
t = 0.
Given:
b
6
m
c
2
s
0.02
1
sp1
2
2000 m
s
Solution:
b c sp
ap
´
µ
¶
vp
vp dvp
0
vp
´
µ
¶
dsp
sp
b c sp dsp
0
2
vp
b sp 2
dsp
vp
t
dvp
dt
µ́
µ
µ
¶
sp
0
c 2
sp
2
2
2b sp c sp
1
2
2b sp c sp
dsp
2
2b sp1 c sp1
vp1
t1
µ́
µ
µ
¶
sp1
0
vp1
1
2
2b sp c sp
dsp
t1
322.49
m
s
19.27 s
Ans.
Ans.
12–14.
The acceleration of a rocket traveling upward is given by ap = b + c sp. Determine the time
needed for the rocket to reach an altitute sp1. Initially, vp = 0 and sp = 0 when t = 0.
Given:
b
6
m
c
2
s
0.02
1
sp1
2
100 m
s
Solution:
b c sp
ap
´
µ
¶
vp
´
µ
¶
vp dvp
0
dvp
dsp
sp
b c sp dsp
0
2
vp
b sp 2
dsp
vp
t
vp
dt
µ́
µ
µ
¶
sp
0
c 2
sp
2
2
2b sp c sp
1
2
2b sp c sp
dsp
2
2b sp1 c sp1
vp1
t1
µ́
µ
µ
¶
sp1
0
vp1
1
2
2b sp c sp
dsp
t1
37.42
m
s
5.62 s
Ans.
12–15.
A train starts from rest at station A and accelerates at
0.5 m>s2 for 60 s. Afterwards it travels with a constant
velocity for 15 min. It then decelerates at 1 m>s2 until it is
brought to rest at station B. Determine the distance
between the stations.
SOLUTION
Kinematics: For stage (1) motion, v0 = 0, s0 = 0, t = 60 s, and ac = 0.5 m>s2. Thus,
+ B
A:
s = s0 + v0t +
s1 = 0 + 0 +
+ B
A:
1 2
at
2 c
1
(0.5)(602) = 900 m
2
v = v0 + act
v1 = 0 + 0.5(60) = 30 m>s
For stage (2) motion, v0 = 30 m>s, s0 = 900 m, ac = 0 and t = 15(60) = 900 s. Thus,
+ B
A:
s = s0 + v0t +
1 2
at
2 c
s2 = 900 + 30(900) + 0 = 27 900 m
For stage (3) motion, v0 = 30 m>s, v = 0, s0 = 27 900 m and ac = - 1 m>s2. Thus,
+ B
A:
v = v0 + act
0 = 30 + ( - 1)t
t = 30 s
+
:
s = s0 + v0t +
1 2
act
2
s3 = 27 900 + 30(30) +
= 28 350 m = 28.4 km
1
( - 1)(302)
2
Ans.
12–16.
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.
SOLUTION
¢s = (sC - sA) = 2 m
sT = (0.5 + 1.5 + 1.5 + 2.5) = 6 m
t = (2 + 4) = 6 s
vavg =
2
¢s
= = 0.333 m>s
t
6
(vsp)avg =
sT
6
= = 1 m>s
t
6
Ans.
Ans.
12–17.
The acceleration of a particle as it moves along a straight
line is given by a = 14t 3 - 12 m>s2, where t is in seconds. If
s = 2 m and v = 5 m>s when t = 0, determine the
particle’s velocity and position when t = 5 s. Also,
determine the total distance the particle travels during this
time period.
SOLUTION
v
L5
t
dv =
L0
3
(4 t - 1) dt
v = t4 - t + 5
s
L2
t
ds =
s =
L0
(t4 - t + 5) dt
1 5
1
t - t2 + 5 t + 2
2
5
When t = 5 s,
v = 625 m>s
Ans.
s = 639.5 m
Ans.
Since v Z 0 then
d = 639.5 - 2 = 637.5 m
Ans.
12–18.
A car can have an acceleration and a deceleration a. If it starts from rest, and can have a
maximum speed v, determine the shortest time it can travel a distance d at which point it stops.
Given:
a
5
m
v
2
60
s
Solution:
t1
a t1
d
§ t1 ·
¨ ¸
¨ t2 ¸
¨ t3 ¸
¨ ¸
¨ d1 ¸
¨d ¸
© 2¹
d
1200 m
Assume that it can reach maximum speed
Guesses
Given
m
s
1s
t2
v
1 2
a t1
2
d2 v t3 t2 Find t1 t2 t3 d1 d2
t3
2s
d1
d1
3s
d2
d2
2m
d1 v t2 t1
1
2
a t3 t2
2
§ t1 ·
¨ ¸
¨ t2 ¸
¨t ¸
© 3¹
1m
0
§ 12 ·
¨ 20 ¸ s
¨ ¸
© 32 ¹
v a t3 t2
§ d1 ·
¨ ¸
© d2 ¹
t3 32 s
§ 360 ·
¨
¸m
© 840 ¹
Ans.
12–19.
A particle travels to the right along a straight line with a
velocity v = [5>14 + s2] m>s, where s is in meters. Determine
its position when t = 6 s if s = 5 m when t = 0.
SOLUTION
5
ds
=
dt
4 + s
s
L5
(4 + s) ds =
t
L0
5 dt
4 s + 0.5 s2 - 32.5 = 5 t
When t = 6 s,
s2 + 8 s - 125 = 0
Solving for the positive root
s = 7.87 m
Ans.
12–20.
A particle is moving along a straight line such that its acceleration is defined as a = k s2. If v =
v0 when s = sp0 and t = 0, determine the particle’s velocity as a function of position.
Given:
k
1
4
100
v0
2
ms
m
s
sp0
10 m
Solution:
d
v v
dsp
a
1
2
v
2
v v0
2
0
1
2
v0 v
´
µ v dv
¶v
2
k sp
3
2
3
3
3
k sp sp0
3
3
k sp sp0
Ans.
sp
´
µ
¶s
p0
2
k sp dsp
12–21.
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 ).
SOLUTION
Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2.
(+ T)
dt =
t
L0
dv
a
v
dv
2
L0 9.81[1 - (0.01v) ]
dt =
v
t =
v
1
dv
dv
c
+
d
9.81 L0 2(1 + 0.01v)
2(1
0.01v)
L0
9.81t = 50ln a
v =
1 + 0.01v
b
1 - 0.01v
100(e0.1962t - 1)
(1)
e0.1962t + 1
a) When t = 5 s, then, from Eq. (1)
v =
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)
vmax = 100 m>s
Ans.
12–22.
A particle moves along a straight line such that its position is defined by s = bt2 + ct + d.
Determine the average velocity, the average speed, and the acceleration of the particle when t = t1.
Given:
b
1
m
2
c
6
s
m
s
d
t0
5m
t1
0 s
6 s
Solution:
2
bt ct d
sp ( t)
Find the critical time
vavevel
vavespeed
a1
a t1
t2
v ( t)
d
sp ( t)
dt
2s
Given
sp t1 sp t0
t1
sp t1 sp t2
a ( t)
v t2
0
d
v ( t)
dt
t2
Find t2
vavevel
sp t2 sp t0
t1
0
vavespeed
a1
2
m
2
s
t2
m
s
3
3s
Ans.
m
s
Ans.
Ans.
12–23.
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 = (g/vf2)(v2f-v2), determine the time needed for the velocity to become v < vf.
Initially the particle falls from rest.
Solution:
dv
dt
a
g
vf
2
v
2
vf v
1
µ́
dv
µ
2
2
v
v
f
µ
¶
2
t
g ´
µ 1 dt
2¶
vf 0
0
1 § vf v ·
ln¨
¸
2vf © vf v ¹
§ g ·t
¨ 2¸
© vf ¹
t
§ vf v ·
¸
2g © vf v ¹
vf
ln ¨
Ans.
12–24.
A particle is moving along a straight line such that its
velocity 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.
SOLUTION
v = - 4s2
ds
= - 4s2
dt
s
L2
s - 2 ds =
t
L0
-4 dt
-s - 1| s2 = - 4t|t0
t =
1 -1
(s - 0.5)
4
s =
2
8t + 1
v = -4a
a =
2
2
16
b = m>s
8t + 1
(8t + 1)2
16(2)(8t + 1)(8)
dv
256
=
=
m>s2
dt
(8t + 1)4
(8t + 1)3
Ans.
Ans.
12–25.
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.
SOLUTION
Velocity: v0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have
A+TB
dv = adt
v
L27
t
dv =
L0
-6tdt
v = A 27 - 3t2 B m>s
(1)
At v = 0, from Eq. (1)
0 = 27 - 3t2
t = 3.00 s
Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result v = 27 - 3t2 and applying
Eq. 12–1, we have
A+TB
ds = vdt
s
L0
t
ds =
L0
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.
12–26.
When two cars A and B are next to one another, they are
traveling in the same direction with speeds vA and vB ,
respectively. If B maintains its constant speed, while A
begins to decelerate at aA , determine the distance d
between the cars at the instant A stops.
A
d
SOLUTION
Motion of car A:
v = v0 + act
0 = vA - aAt
t =
vA
aA
v2 = v20 + 2ac(s - s0)
0 = v2A + 2( - aA)(sA - 0)
sA =
v2A
2aA
Motion of car B:
sB = vBt = vB a
vA
vAvB
b =
aA
aA
The distance between cars A and B is
sBA = |sB - sA| = `
v2A
vAvB
2vAvB - v2A
` = `
`
aA
2aA
2aA
B
Ans.
12–27.
A particle is moving along a straight line such that when it
is at the origin it has a velocity of 4 m>s. If it begins to
decelerate at the rate of a = 1- 1.5v1>22 m>s2, where v is in
m>s, determine the distance it travels before it stops.
SOLUTION
1
dv
= - 1.5v2
dt
a =
v
L4
1
t
1
v- 2 dv =
v
L0
- 1.5 dt
t
2v2 4 = - 1.5t 0
1
2a v2 - 2 b = - 1.5t
v = (2 - 0.75t)2 m>s
s
L0
ds =
(1)
t
L0
t
(2 - 0.75t)2 dt =
L0
(4 - 3t + 0.5625t2) dt
s = 4t - 1.5t2 + 0.1875t3
(2)
From Eq. (1), the particle will stop when
0 = (2 - 0.75t)2
t = 2.667 s
s|t = 2.667 = 4(2.667) - 1.5(2.667)2 + 0.1875(2.667)3 = 3.56 m
Ans.
12–28.
A particle travels to the right along a straight line with a
velocity v = [5>14 + s2] m>s, where s is in meters.
Determine its deceleration when s = 2 m.
SOLUTION
v =
5
4+s
v dv = a ds
dv =
-5 ds
(4 + s)2
5
- 5 ds
b = a ds
a
(4 + s) (4 + s)2
a =
- 25
(4 + s)3
When s = 2 m
a = -0.116 m>s2
Ans.
12–29.
A particle moves along a straight line with an acceleration
1>2
a = 2v m>s2, where v is in m>s. If s = 0, v = 4 m>s when
t = 0, determine the time for the particle to achieve a
velocity of 20 m>s. Also, find the displacement of particle
when t = 2 s.
SOLUTION
Velocity:
+ B
A:
dt =
dv
a
t
L0
v
dt =
dv
L0 2v1>2
t
v
t ƒ 0 = v1>2 ƒ 4
t = v1>2 - 2
v = (t + 222
When v = 20 m>s,
20 = 1t + 222
t = 2.47 s
Ans.
Position:
+ B
A:
ds = v dt
s
t
ds =
L0
s
s2 =
0
L0
(t + 2)2 dt
t
1
(t + 2)3 2
3
0
s =
1
[(t + 2)3 - 23]
3
=
1 2
t(t + 6t + 12)
3
When t = 2 s,
s =
1
(2)[(2)2 + 6(2) + 12]
3
= 18.7 m
Ans.
12–30.
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.
SOLUTION
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 kilometer,
ac = 0.048 m>s2. Thus,
+ B
A:
v0 = 10 m>s,
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 ac = 0.048 m>s.2 Thus,
+ B
A:
v = v0 + act
14 = 2 + 0.048t
t = 250 s
Ans.
12–31.
The acceleration of a particle along a straight line is defined
by a = 12t - 92 m>s2, where t is in seconds. At t = 0,
s = 1 m and v = 10 m>s. When t = 9 s, determine (a) the
particle’s position, (b) the total distance traveled, and
(c) the velocity.
SOLUTION
a = 2t - 9
v
L10
t
dv =
L0
12t - 92 dt
v - 10 = t2 - 9 t
v = t2 - 9 t + 10
s
L1
t
ds =
s-1 =
s =
L0
1t2 - 9t + 102 dt
13
t - 4.5 t2 + 10 t
3
13
t - 4.5 t2 + 10 t + 1
3
Note when v = t2 - 9 t + 10 = 0:
t = 1.298 s and t = 7.701 s
When t = 1.298 s,
s = 7.13 m
When t = 7.701 s,
s = - 36.63 m
When t = 9 s,
s = - 30.50 m
(a)
s = - 30.5 m
(b)
sTo t = ( 7.13 - 1) + 7.13 + 36.63 + (36.63 - 30.50)
(c)
Ans.
sTo t = 56.0 m
Ans.
v = 10 m>s
Ans.
12–32.
The acceleration of a particle traveling along a straight line
1 1>2
is a = s m>s2, where s is in meters. If v = 0, s = 1 m
4
when t = 0, determine the particle’s velocity at s = 2 m.
SOLUTION
Velocity:
+ B
A:
v dv = a ds
v
L0
s
v dv =
1 1>2
s ds
L1 4
v
s
v2 2
1
= s3>2 `
2 0
6
1
v =
1
23
When s = 2 m, v = 0.781 m>s.
1s3>2 - 121>2 m>s
Ans.
12–33.
At t = 0 bullet A is fired vertically with an initial (muzzle)
velocity of 450 m/s. When t = 3 s, bullet B is fired upward
with a muzzle velocity of 600 m/s. Determine the time t, after
A is fired, as to when bullet B passes bullet A. At what
altitude does this occur?
SOLUTION
+ c sA = (sA)0 + (vA)0 t +
sA = 0 + 450 t +
1 2
a t
2 c
1
(- 9.81) t2
2
+ c sB = (sB)0 + (vB)0 t +
sB = 0 + 600(t - 3) +
1
a t2
2 c
1
( -9.81)(t - 3)2
2
Require sA = sB
450 t - 4.905 t2 = 600 t - 1800 - 4.905 t2 + 29.43 t - 44.145
t = 10.3 s
Ans.
h = sA = sB = 4.11 km
Ans.
12–34.
A boy throws a ball straight up from the top of a 12-m high
tower. If the ball falls past him 0.75 s later, determine the
velocity at which it was thrown, the velocity of the ball when
it strikes the ground, and the time of flight.
SOLUTION
Kinematics: When the ball passes the boy, the displacement of the ball in equal to zero.
Thus, s = 0. Also, s0 = 0, v0 = v1, t = 0.75 s, and ac = - 9.81 m>s2.
A+cB
s = s0 + v0t +
1 2
at
2 c
0 = 0 + v110.752 +
1
1-9.81210.7522
2
v1 = 3.679 m>s = 3.68 m>s
Ans.
When the ball strikes the ground, its displacement from the roof top is s = - 12 m.
Also, v0 = v1 = 3.679 m>s, t = t2, v = v2, and ac = - 9.81 m>s2.
A+cB
s = s0 + v0t +
1 2
at
2 c
- 12 = 0 + 3.679t2 +
1
1- 9.812t22
2
4.905t22 - 3.679t2 - 12 = 0
t2 =
3.679 ; 21 - 3.67922 - 414.90521 -122
214.9052
Choosing the positive root, we have
t2 = 1.983 s = 1.98 s
Ans.
Using this result,
A+cB
v = v0 + act
v2 = 3.679 + 1 -9.81211.9832
= - 15.8 m>s = 15.8 m>s T
Ans.
12–35.
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.
SOLUTION
g
dv
= a = ¢ 2 ≤ A v2f - v2 B
dt
vf
v
dy
L0 v2f - v2¿
=
t
g
v2f L0
dt
vf + v y
g
1
ln ¢
≤` = 2t
2vf
vf - v 0
vf
t =
t =
vf
2g
vf
2g
ln ¢
ln ¢
t = 0.549 a
vf + v
vf - v
≤
vf + vf> 2
vf - vf> 2
vf
g
b
≤
Ans.
12–36.
A particle is moving with a velocity of v0 when s = 0 and
t = 0. If it is subjected to a deceleration of a = - kv3, where
k is a constant, determine its velocity and position as
functions of time.
SOLUTION
a =
v
Lv0
v-3 dv =
dv
= - ky3
dt
t
-k dt
L0
1
- 1v-2 - v0-22 = - kt
2
1
-2
1
v = ¢ 2kt + ¢ 2 ≤ ≤
v0
Ans.
ds = v dt
s
L0
t
ds =
L0
dt
1
¢ 2kt + ¢
2
1
≤≤
v20
1
s =
2
1
2 ¢ 2kt + ¢ 2 ≤ ≤
v0
2k
t
4
0
1
s =
2
1
1
1
C ¢ 2kt + ¢ 2 ≤ ≤ S
v0
k
v0
Ans.
12–37.
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.
SOLUTION
v dv = a dy
q
0
Ly
v dv = - g0R
2
dy
L0 (R + y)
2
g 0 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.
12–38.
Accounting for the variation of gravitational acceleration
a with respect to altitude y (see Prob. 12–37), 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. 12–37.
SOLUTION
From Prob. 12–37,
(+ c )
a = -g0
R2
(R + y)2
Since a dy = v dv
then
y
-g0 R2
v
dy
Ly0 (R + y)
2
=
L0
v dv
g0 R2 c
y
1
v2
d =
R + y y0
2
g0 R2[
1
1
v2
] =
R + y
R + y0
2
Thus
v = -R
2g0 (y0 - y)
A (R + y)(R + y0)
When y0 = 500 km,
v = -6356(103)
Ans.
y = 0,
2(9.81)(500)(103)
A 6356(6356 + 500)(106)
v = - 3016 m>s = 3.02 km>s T
Ans.
12–39.
A car starting from rest moves along a straight track with an acceleration as shown. Determine
the time t for the car to reach speed v.
Given:
v
50
a1
8
m
s
m
2
s
t1
10 s
Solution:
Assume that t > t1
t
Guess
12 s
2
Given
v
a1 t 1
t1 2
a1 t t 1
t
Find ( t)
t
11.25 s
Ans.
12–40.
If the position of a particle is defined by sp = b sin(ct) + d, construct the s-t, v-t, and a-t graphs
for 0 d t d T.
Given:
b
3m
c
( b sin ( c t) d)
1
m
S 1
d
4 s
T
8m
10 s
t
0 0.01T T
Solution:
sp ( t)
vp ( t)
b c cos ( c t)
s
m
ap ( t )
2
b c sin ( c t)
s
m
2
Distance in m
15
sp(t) 10
5
0
2
4
6
t
Time in Seconds
8
10
4
8
10
8
10
Velocity in m/s
4
2
vp(t)
0
–2
–4
0
2
6
t
Time in Seconds
Acceleration in m/sˆ2
2
ap(t)
0
–2
0
2
6
t
Time in Seconds
4
12–41.
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.
SOLUTION
For stage (1) motion,
+ B
A:
v1 = v0 + 1ac21t
vmax = 0 + 1ac21t1
vmax = 1ac21t1
+ B
A:
(1)
v12 = v02 + 21ac211s1 - s02
vmax 2 = 0 + 21ac2111000 - 02
1ac21 =
vmax 2
2000
(2)
Eliminating 1ac21 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 = 1t2 - t12. Thus,
+ B
A:
s2 = s1 + v1t +
1
1a 2 t2
2 c 2
1000 + 2000 = 1000 + vmax1t2 - t12 + 0
t2 - t1 =
2000
vmax
(4)
For stage (3) motion, the train travels for t = 360 - t2. Thus,
+ B
A:
v3 = v2 + 1ac23t
0 = vmax - 1ac231360 - t22
vmax = 1ac231360 - t22
+ B
A:
(5)
v32 = v22 + 21ac231s3 - s22
0 = vmax 2 + 23 - 1ac23414000 - 30002
1ac23 =
vmax 2
2000
(6)
Eliminating 1ac23 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
Based on the above results, the v-t graph is shown in Fig. a.
v = vmax for 2 min < t < 4 min.
Ans.
12–42.
A particle starts from s = 0 and travels along a straight line
with a velocity v = (t2 - 4t + 3) m > s, where t is in
seconds. Construct the v -t and a- t graphs for the time
interval 0 … t … 4 s.
SOLUTION
a–t Graph:
a =
d 2
dv
=
1t - 4t + 32
dt
dt
a = (2t - 4) m>s2
Thus,
a|t = 0 = 2(0) - 4 = - 4 m>s2
a|t = 2 s = 0
a|t = 4 s = 2(4) - 4 = 4 m>s2
The a - t graph is shown in Fig. a.
v – t Graph: The slope of the v - t graph is zero when a =
a = 2t - 4 = 0
dv
= 0. Thus,
dt
t = 2s
The velocity of the particle at t = 0 s, 2 s, and 4 s are
v|t = 0 s = 02 - 4(0) + 3 = 3 m>s
v|t = 2 s = 22 - 4(2) + 3 = - 1 m>s
v|t = 4 s = 42 - 4(4) + 3 = 3 m>s
The v- t graph is shown in Fig. b.
12–43.
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.
SOLUTION
12–44. 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
DT
6T
A Dv
6T
dt
6
v 0 at 0 3T2
2
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
12–45. The snowmobile moves along a straight course
according to the W–t graph. Construct the s–t and a–t graphs
for the same 50-s time interval. When U 0, T 0.
U V Graph: The position function in terms of time t can be obtained by applying
ET
2
12
y U 3 U 4 ms.
. For time interval 0 s Š U 30 s, y EU
30
5
v (m/s)
12
ET yEU
T
(0
U
ET 2
UEU
(0 5
1
T 3 U2 4 m
5
At U 30 s ,
T 1
302 180 m
5
For time interval 30 sV • 50 s,
ET yEU
T
U
ET (180 m
12EU
(30 T
T (12U 180) m
At U 50 s,
T 12(50) 180 420 m
C V Graph: The acceleration function in terms of time t can be obtained by applying
Ey
2
Ey
B . For time interval 0 s • V30 s and 30 sV • 50 s, B EU
EU
5
Ey
2
0.4 ms and B 0, respectively.
EU
t (s)
30
50
12–46.
The velocity of a car is plotted as shown. Determine the
total distance the car moves until it stops 1t = 80 s2.
Construct the a–t graph.
v (m/s)
10
SOLUTION
t (s)
Distance Traveled: The total distance traveled can be obtained by computing the
area under the v - t graph.
s = 10(40) +
1
(10)(80 - 40) = 600 m
2
Ans.
dv
a – t Graph: The acceleration in terms of time t can be obtained by applying a =
.
dt
For time interval 0 s … t 6 40 s,
a =
For time interval 40 s 6 t … 80 s,
a =
dv
= 0
dt
v - 10
0 - 10
1
, v = a - t + 20 b m>s.
=
t - 40
80 - 40
4
dv
1
= - = - 0.250 m s2
dt
4
For 0 … t 6 40 s, a = 0.
For 40 s 6 t … 80, a = - 0.250 m s2 .
40
80
12–47.
The v–s graph for a go-cart traveling on a straight road is
shown. Determine the acceleration of the go-cart at
s = 50 m and s = 150 m. Draw the a–s graph.
v (m/s)
8
SOLUTION
For 0 … s 6 100
v = 0.08 s,
100
dv = 0.08 ds
a ds = (0.08 s)(0.08 ds)
a = 6.4(10 - 3) s
At s = 50 m,
a = 0.32 m>s2
Ans.
For 100 6 s 6 200
v = - 0.08 s + 16,
dv = - 0.08 ds
a ds = (-0.08 s + 16)( - 0.08 ds)
a = 0.08(0.08 s - 16)
At s = 150 m,
a = - 0.32 m>s2
Ans.
Also,
v dv = a ds
a = v(
dv
)
ds
At s = 50 m,
a = 4(
8
) = 0.32 m>s2
100
Ans.
At s = 150 m,
a = 4(
-8
) = -0.32 m>s2
100
At s = 100 m, a changes from amax = 0.64 m>s2
to amin = -0.64 m>s2 .
Ans.
200
s (m)
12–48.
The v–t graph for a particle moving through an electric field
from one plate to another has the shape shown in the figure.
The acceleration and deceleration that occur are constant
and both have a magnitude of 4 m>s2. If the plates are
spaced 200 mm apart, determine the maximum velocity vmax
and the time t¿ for the particle to travel from one plate to
the other. Also draw the s–t graph. When t = t¿>2 the
particle is at s = 100 mm.
smax
v
s
vmax
SOLUTION
ac = 4 m/s2
t¿/ 2
s
= 100 mm = 0.1 m
2
v2 = v20 + 2 ac(s - s0)
v2max = 0 + 2(4)(0.1 - 0)
vmax = 0.89442 m>s
= 0.894 m>s
Ans.
v = v0 + ac t¿
t¿
0.89442 = 0 + 4( )
2
t¿ = 0.44721 s
= 0.447 s
s = s0 + v0 t +
s = 0 + 0 +
1
a t2
2 c
1
(4)(t)2
2
s = 2 t2
When t =
0.44721
= 0.2236 = 0.224 s,
2
s = 0.1 m
v
t
ds = -
L0.894
4 dt
L0.2235
v = - 4 t +1.788
s
t
ds =
L0.1
1 -4t + 1.7882 dt
L0.2235
s = - 2 t2 + 1.788 t - 0.2
When t = 0.447 s,
s = 0.2 m
Ans.
t¿
t
12–49.
The v–t graph for a particle moving through an electric field
from one plate to another has the shape shown in the figure,
where t¿ = 0.2 s and vmax = 10 m>s. Draw the s–t and a–t graphs
for the particle. When t = t¿>2 the particle is at s = 0.5 m.
smax
v
s
vmax
SOLUTION
For 0 6 t 6 0.1 s,
t¿/ 2
v = 100 t
dv
= 100
dt
a =
ds = v dt
s
L0
t
ds =
L0
100 t dt
s = 50 t 2
When t = 0.1 s,
s = 0.5 m
For 0.1 s 6 t 6 0.2 s,
v = -100 t + 20
a =
dv
= - 100
dt
ds = v dt
s
L0.5
t
ds =
1 -100t + 202dt
L0.1
s - 0.5 = ( -50 t 2 + 20 t - 1.5)
s = - 50 t 2 + 20 t - 1
When t = 0.2 s,
s = 1m
When t = 0.1 s, s = 0.5 m and a changes from 100 m/s2
to -100 m/s2. When t = 0.2 s, s = 1 m.
t¿
t
12–50.
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)
SOLUTION
20
0 … t … 5
20
¢v
=
= 4 m>s2
¢t
5
a =
5 … t … 20
a =
20 … t … 30
a =
5
¢v
20 - 20
= 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
s
0 … t … 5s
v = 4t;
ds = v dt;
L0
t
ds =
L0
4t dt;
s
5 … t … 20 s
v = 20;
ds = v dt;
L50
ds =
t
v = 2(30 - t);
ds = v dt;
For 0 … t 6 5 s, a = 4 m>s2.
For 20 s 6 t … 30 s, a = - 2 m>s2.
At t = 5 s, s = 50 m. At t = 20 s, s = 350 m.
At t = 30 s, s = 450 m.
s = 20t - 50
20 dt;
L5
s
20 … t … 30 s
s = 2t2
t
ds =
L350
L20
2(30 - t) dt;
s = -t2 + 60t - 450
20
30
t (s)
12–51.
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)
a
0.1t
SOLUTION
v t Graph: For the time interval 0 … t 6 30 s, the initial condition is v = 0 when
t = 0 s.
+ B
A:
dv = adt
v
L0
t
dv =
L0
0.1tdt
v = A 0.05t2 B m>s
When t = 30 s,
v t = 30 s = 0.05 A 302 B = 45 m>s
or 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
dv =
L30 s
L45 m>s
v = ¢-
¢-
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
This equation is the same as the one obtained previously.
The slope of the v–tgraph is zero when t = 75 s, which is the instant a =
v t = 75 s = -
1
A 752 B + 5(75) - 75 = 112.5 m>s
30
dv
= 0.Thus,
dt
1
a
3
( 15 )t
5
t¿
30
75
t(s)
12–51. continued
The v–t graph is shown in Fig. a.
s t 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
s
L0
t
ds =
s = a
L0
0.05t2 dt
1 3
t bm
60
When t = 30 s,
1
A 303 B = 450 m
60
s 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
t
ds =
L450 m
L30 s
a-
1 2
t + 5t - 75 b dt
30
5
1
s = a - t3 + t2 - 75t + 750b m
90
2
When t = 75 s and t¿ = 133.09 s,
s t = 75 s = -
1
5
A 753 B + A 752 B - 75(75) + 750 = 4500 m
90
2
s t = 133.09 s = -
5
1
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.
When t = 30 s,
v = 45 m/s and s = 450 m.
When t = 75 s,
v = vmax = 112.5 m/s and s = 4500 m.
When t = 133 s,
v = 0 and s = 8857 m.
Ans.
12–52.
The snowmobile moves along a straight course according to
the v –t graph. Construct the s–t and a–t graphs for the same
100-s time interval. When t = 0, s = 0.
v (m/s)
10
SOLUTION
s t Graph: The position function in terms of time t can be obtained by applying
10
ds
1
v =
t = a t b m>s.
. For time interval 0 s … t 6 60 s, v =
60
6
dt
t (s)
60
100
ds = vdt
s
L0
t
ds =
s = a
At t = 60 s ,
1
tdt
L0 6
1 2
tbm
12
700
S = 10 t – 3 0 0
300
ଵ
S = ଵଶ t
1
s =
A 602 B = 300 m
12
2
60
100
60
100
For time interval 60 s<t ◊ 100 s,
ds = vdt
s
t
ds =
10dt
L60 s
L300 m
s = (10t - 300) m
At t = 100 s,
s = 10(100) - 300 = 700 m
a t Graph: The acceleration function in terms of time t can be obtained by applying
10
dv
dv
. For time interval 0 s ◊ t<60 s and 60 s<t ◊ 100 s, a =
a =
=
60
dt
dt
dv
2
1
=
m>s and a =
= 0, respectively.
dt
6
ଵ
଺
12–53.
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.
a (m/s2)
SOLUTION
25
Since v =
v–t graph.
L
B
18
a dt, the constant lines of the a–t graph become sloping lines for the
At t = 15 s,
v = (18)(15) = 270 m>s
At t = 20 s,
v = 270 + (25)(20 - 15) = 395 m>s
s–t graph.
L
t (s)
15
The numerical values for each point are calculated from the total area under the
a–t graph to the point.
Since s =
A
v dt, the sloping lines of the v–t graph become parabolic curves for the
The numerical values for each point are calculated from the total area under the v–t
graph to the point.
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 m>s 2
v = v0 + ac t = 0 + 18t
s = s0 + v0 t +
1
ac t2 = 0 + 0 + 9t2
2
When t = 15:
v = 18(15) = 270 m>s
s = 9(15)2 = 2025 m = 2.025 km
15 … t … 20:
a = 25 m>s 2
v = v0 + ac t = 270 + 25(t - 15)
s = s0 + v0 t +
1
1
ac t2 = 2025 + 270(t - 15) + (25)(t - 15)2
2
2
When t = 20:
v = 395 m>s
s = 3687.5 m = 3.69 km
20
12–54
A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by
the v-t graph. Determine the motorcycle's acceleration and position when t = t4 and t = t5.
Given:
m
s
v0
5
t1
4s
t2
10 s
t3
15 s
t4
8s
t5
12 s
Solution:
At t
t4
s4
At t
t5
a4
dv
dt
0
Ans.
1
v0 t1 t4 t1 v0
2
s4
30.00 m
Ans.
Because t2 t5 t3 then
a5
s5
Because t1 t4 t2 then
Ans.
v0
a5
t3 t2
1
1
1 t3 t1 v0 v0 t2 t1 v0 t3 t2 2
2
2 t3 1.00
m
2
Ans.
s
t5
t2
v0 t3 t5
s5
48.00 m
Ans.
12–55.
A race car starting from rest travels along a straight
road and for 10 s has the acceleration shown. Construct
the v–t graph that describes the motion and find the
distance traveled in 10 s.
a (m/s2)
6
6
a=
SOLUTION
v -t Graph: The velocity function in terms of time t can be obtained by applying
dv
formula a =
. For time interval 0 s … t 6 6 s,
dt
dv = adt
v
t
L0
v = a
At t = 6 s,
v =
1 2
t dt
L0 6
dv =
1 3
t b m>s
18
1
A 63 B = 12.0 m>s,
18
For time interval 6 s 6 t … 10 s,
dv = adt
v
t
L12.0m>s
dv =
L6s
6dt
v = (6t - 24) m>s
At t = 10 s,
v = 6(10) - 24 = 36.0 m>s
ds
Position: The position in terms of time t can be obtained by applying v =
.
dt
For time interval 0 s … t 6 6 s,
ds = vdt
s
L0
t
ds =
s = ¢
1 3
t dt
L0 18
1 4
t ≤m
72
1
A 64 B = 18.0 m.
72
When t = 6 s, v = 12.0 m>s and s =
For time interval 6 s 6 t … 10 s,
ds = vdt
s
t
dv =
L18.0 m
L6s
(6t - 24)dt
s = A 3t2 - 24t + 54 B m
When t = 10 s, v = 36.0 m>s and s = 3 102 - 24(10) + 54 = 114 m
Ans.
1
—
6
t2
t (s)
6
10
A car travels up a hill with the speed shown.
12–56.
Determine the total distance the car travels until it stops
(U 60 s). Plot the B U graph.
v (m/s)
10
t (s)
Distance traveled is area under v U graph.
T (10)(30)
1
(10)(30) 450 m
2
30
Ans.
60
12–5 7. 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 T–U
and B–U graphs.
v(m/s)
30
v 0.5t 45
vt
U V Graph: For the time interval 0 Š U 30 s, the initial condition is T 0 when
U 0 s.
ET vEU
T
(0
U
ET UEU
(0
U2
T A Im
2
When U 30 s,
T 302
450 m
2
For the time interval 30 s U Š 90 s, the initial condition is T 450 m when
U 30 s.
ET vEU
T
U
ET (30s
(450 m
1
T 3 U2
4
( 0.5U
45)EU
45U 675 4 m
When U 90 s,
T U 90 s 1
902 4
45(90) 675 1350 m
Ans.
The T U graph shown is in Fig. a.
CV Graph: For the time interval 0 U 30 s,
B Ev
E
(U) 1 ms2
EU
EU
For the time interval 30 s U Š 90 s,
B Ev
E
( 0.5U
EU
EU
45) 0.5 ms2
The B U graph is shown in Fig. b.
Note: Since the change in position of the car is equal to the area under the v U
graph, the total distance traveled by the car is
cT (
vEU
T U 90 s 0 1
(90)(30)
2
T U 90 s 1350 s
t(s)
30
60
90
12–58.
The jet-powered boat starts from rest at s = 0 and travels
along a straight line with the speed described by the graph.
Construct the s -t and a- t graph for the time interval
0 … t … 50 s.
v (m/s)
v ⫽ 4.8 (10 ⫺3)t 3
75
v ⫽ ⫺3t ⫹ 150
SOLUTION
s–t Graph: The initial condition is s = 0 when t = 0.
+ B
A:
ds = vdt
s
L0
t
ds =
L0
4.8(10-3)t3 dt
s = [1.2110-3)t4 ]m
At t = 25 s,
s|t = 25 s = 1.2(10-3)(254) = 468.75 m
For the time interval 25 s 6 t … 50 s, the initial condition s = 468.75 m when
t = 25 s will be used.
+ B
A:
ds = vdt
s
t
ds =
L468.75 m
L25 s
( -3t + 150) dt
3
s = a - t2 + 150t - 2343.75 b m
2
When t = 50 s,
3
s|t = 50 s = - (502) + 150(50) - 2343.75 = 1406.25 m
2
The s - t graph is shown in Fig. a.
a–t Graph: For the time interval 0 … t 6 25 s,
a =
dv
d
= [4.8(10-3)t3] = (0.0144t2) m > s2
dt
dt
When t = 25 s,
a|t = 25 s = 0.0144(252) m > s2 = 9 m > s2
For the time interval 25 s 6 t … 50 s,
a =
dv
d
= (- 3t + 150) = - 3 m > s2
dt
dt
The a -t graph is shown in Fig. b.
When t = 25 s,
a = amax = 9 m > s2 and s = 469 m.
When t = 50 s,
s = 1406 m.
t (s)
25
50
12–59
From experimental data, the motion of a jet plane while traveling along a runway is defined by
the v–t graph shown. Construct the s-t and a-t graphs for the motion.
Given:
m
s
v1
80
t1
10 s
t2
40 s
Solution:
k1
v1
t1
k2
0
m
2
s
2
W1
0 0.01t1 t1
s1 W 1
§ 1 k W 2· m
¨ 1 1¸
©2
¹
a1 W 1
k1
s
m
W2
t1 1.01t1 t2
s2 W 2
§ v W 1 k t 2· m
¨1 2
1 1 ¸
2
©
¹
a2 W 2
k2
s
m
35
40
2
Position (m)
3000
s1 W 1
2000
s2 W 2
1000
0
0
5
10
15
20
W1 W2
Time (s)
25
30
Acceleration (m/s^2)
10
a1 W 1
5
a2 W 2
0
0
5
10
15
20
W1 W2
Time (s)
25
30
35
40
12–60.
A car travels along a straight road with the speed shown by
the v - t graph. Plot the a- t graph.
v (m/s)
6
1
v⫽—
5 t
1
v ⫽ ⫺—
3 (t ⫺ 48)
SOLUTION
a–t Graph: For 0 … t 6 30 s,
v =
1
t
5
a =
dv
1
= = 0.2 m > s2
dt
5
For 30 s 6 t … 48 s
1
v = - (t - 48)
3
dv
1
a =
= - (1) = - 0.333 m > s2
dt
3
Using these results, a- t graph shown in Fig. a can be plotted.
t (s)
30
48
12–61.
A car travels along a straight road with the speed shown by
the v–t graph. Determine the total distance the car travels
until it stops when t = 48 s. Also plot the s–t and a–t graphs.
v (m/s)
6
v
1
—
5 t
v
1
—
3 (t
48)
SOLUTION
For 0 … t … 30 s,
t (s)
30
1
v = t
5
a =
dv
1
=
dt
5
ds = v dt
s
t
ds =
L0
s =
When t = 30 s,
1
t dt
L0 5
1 2
t
10
s = 90 m,
1
v = - (t - 48)
3
a =
dv
1
= dt
3
ds = v dt
s
L90
t
ds =
1
- 1t - 482dt
3
L30
1
s = - t2 + 16t - 240
6
When t = 48 s,
s = 144 m
Ans.
Also, from the v–t graph
¢s =
L
v dt s - 0 =
1
1621482 = 144 m
2
Ans.
48
12–62
The a-s graph for a train traveling along a straight track is given for 0 d s d s2. Plot the v-s
graph. v = 0 at s = 0.
Given:
s1
200 m
s2
400 m
a1
2
m
2
s
Solution:
V1
0 0.01s1 s1
V2
s1 1.01s1 s2
For 0 s s1
a
ks
v
s1
v
´
µ v dv
¶0
dv
ds
For s1 s s2
2
a1
k
ac2
ac1
ks
s
2
´
µ k s ds
¶0
a1
a
v
2
ks
k 2
s
2
v
´
µ v dv
¶v
dv
ds
v
v1 V 1
1
k V1
v2 V 2
a1 s s1
s
´
µ a1 ds
¶s
2
v1
v
2
2
s
m
2a1 V 2 s1 k s1
1
2 s
m
Velocity in m/s
40
v1 V 1
v2 V 2
20
0
0
50
100
150
200
V1 V2
Distance in m
250
300
350
400
12–63.
The speed of a train during the first minute has been
recorded as follows:
t (s)
0 20
40
60
0 16
21
24
v (m>s)
Plot the v-t graph, approximating the curve as straight-line
segments between the given points. Determine the total
distance traveled.
SOLUTION
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
Ans.
12–64
The v–s graph for a test vehicle is shown. Determine its acceleration at s = s3 and s4.
Given:
m
s
v1
50
s1
150 m
s2
200 m
s3
100 m
s4
175 m
Solution:
a3
a4
§ s3 · § v1 ·
¨ ¸ v1¨ ¸
© s1 ¹ © s1 ¹
a3
§ s2 s4 · § 0 v1 ·
¨
¸ v1¨
¸
© s2 s1 ¹ © s2 s1 ¹
a4
11.11
m
2
Ans.
s
25
m
2
s
Ans.
12–65.
Two cars start from rest side by side and travel along a
straight road. Car A accelerates at 4 m>s2 for 10 s and then
maintains a constant speed. Car B accelerates at 5 m>s2
until reaching a constant speed of 25 m/s and then
maintains this speed. Construct the a–t, v–t, and s–t graphs
for each car until t = 15 s. What is the distance between the
two cars when t = 15 s?
SOLUTION
Car A:
v = v0 + ac t
vA = 0 + 4t
At t = 10 s,
vA = 40 m>s
s = s0 + v0t +
sA = 0 + 0 +
1 2
at
2 c
1
(4)t2 = 2t2
2
At t = 10 s,
sA = 200 m
t 7 10 s,
ds = v dt
sA
L200
t
ds =
L10
40 dt
sA = 40t - 200
At t = 15 s,
sA = 400 m
Car B:
v = v0 + a c t
vB = 0 + 5t
When vB = 25 m/s,
t =
25
= 5s
5
s = s0 + v0t +
sB = 0 + 0 +
1 2
at
2 c
1
(5)t2 = 2.5t2
2
When t = 10 s, vA = (vA)max = 40 m/s and sA = 200 m.
When t = 5 s, sB = 62.5 m.
When t = 15 s, sA = 400 m and sB = 312.5 m.
12–65. continued
At t = 5 s,
sB = 62.5 m
t 7 5 s,
ds = v dt
sB
L62.5
t
ds =
L5
25 dt
sB - 62.5 = 25t - 125
sB = 25t - 62.5
When t = 15 s,
sB = 312.5
Distance between the cars is
¢s = sA - sB = 400 - 312.5 = 87.5 m
Car A is ahead of car B.
Ans.
12–66.
A two-stage rocket is fired vertically from rest at s = 0 with
an acceleration as shown. After 30 s the first stage A burns
out and the second stage B ignites. Plot the v–t and s–t
graphs which describe the motion of the second stage for
0 … t … 60 s.
a (m/s2)
B
A
15
9
SOLUTION
a
0.01t2
For 0 … t … 30 s
v
l
dv =
L0
0.01 t2 dt
L0
v = 0.00333t3
When t = 30 s, v = 90 m>s
For 30 s … t … 60 s
v
L90
l
dv =
L30
15dt
v = 15t - 360
When t = 60 s,
v = 540 m>s
30
60
t (s)
12–67.
A two-stage rocket is fired vertically from rest at s = 0 with
an acceleration as shown. After 30 s the first stage A burns
out and the second stage B ignites. Plot the s–t graph which
describes the motion of the second stage for 0 … t … 60 s.
a (m/s2)
B
A
15
9
a ⫽ 0.01t 2
SOLUTION
v – t Graph: When t = 0, v = 0. For 0 … t … 30 s,
A+cB
30
dv = a dt
v
L0
v2
t
dv =
v
0.01t2dt
L0
0.01 3 t
t 2
3
0
=
0
v = {0.003333t3} m>s
When t = 30 s, v = 0.003333(303) = 90 m>s
For 30 s 6 t … 60 s,
A+cB
dv = a dt
v
t
L90 m>s
v2
dv =
v
L30 s
= 15t 2
90 m>s
15 dt
t
30 s
v - 90 = 15t - 450
v = {15t - 360} m>s
When t = 60 s, v = 15(60) - 360 = 540 m>s
s–t Graph: When t = 0, s = 0. For 0 … t … 30 s,
A+cB
ds = vdt
s
t
ds =
L0
L0
0.003333t3dt
s
s 2 = 0.0008333t4 2
0
t
0
s = {0.0008333 t } m
4
When t = 30 s, s = 0.0008333(304) = 675 m
For 30 s 6 t … 60 s,
A+cB
ds = vdt
s
L675 m
t
ds =
L30 s
(15t - 360) dt
60
t (s)
12–67. continued
s2
s
675 m
= (7.5t2 - 360t) 2
t
30 s
s - 675 = (7.5t2 - 360t) - [7.5(302) - 360(30)]
s = {7.5t2 - 360t + 4725} m
When t = 60 s, s = 7.5(602) - 360(60) + 4725 = 10 125 m
Using these results, the s–t graph shown in Fig. a can be plotted.
12–68.
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
SOLUTION
a s 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
v s Graph: The function of velocity v in terms of s can be obtained by applying
vdv = ads. For the interval 0 m ◊ s<200 m,
vdv = ads
v
L0
s
vdv =
L0
0.01sds
v = (0.1s) m>s
At s = 200 m,
v = 0.100(200) = 20.0 m>s
For the interval 200 m 6 s … 300 m,
vdv = ads
v
s
L20.0 m>s
v =
At s = 300 m,
vdv =
L200 m
( - 0.02s + 6)ds
A 2 - 0.02s2 + 12s - 1200 B m>s
v = 2 - 0.02(3002) + 12(300) - 1200 = 24.5 m>s
200
300
s (m)
12–69.
The s–t graph for a train has been experimentally determined. From the data, construct the v–t
and a–t graphs for the motion; 0 d t d t2. For 0 d t d t1 , the curve is a parabola, and then it
becomes straight for t t t1.
Given:
t1
30 s
t2
40 s
s1
360 m
s2
600 m
Solution:
k1
s1
k2
2
t1
W1
s2 s1
t2 t1
W2
0 0.01t1 t1
2
sp1 ( t)
k1 t
sp2 ( t)
sp1 t1 k2 t t1
t1 1.01t1 t2
v1 ( t)
2 k1 t
a1 ( t )
2 k1
v2 ( t)
k2
a2 ( t )
0
Velocity (m/s)
30
v1 W 1 20
v2 W 2
10
0
0
5
10
15
20
W 1 W 2
Time (s)
25
30
35
40
Acceleration (m/s^2)
1
a 1 W 1 0.5
a2 W 2
0
0
5
10
15
20
W1 W2
Time (s)
25
30
35
40
12–70.
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
SOLUTION
s t Graph: For 0 … s 6 100 m, the initial condition is s = 0 when t = 0 s.
+ B
A:
v
ds
dt =
v
t
v2
4s
s
ds
dt =
L0 2s1>2
L0
t = s1>2
20
s (m)
100
s = A t2 B m
When s = 100 m,
100 = t2
t = 10 s
For 100 m 6 s … 400 m, the initial condition is s = 100 m when t = 10 s.
+ B
A:
0.2s
ds
v
dt =
t
s
dt =
ds
L10 s
L100 m 0.2s
s
t - 10 = 5ln
100
s
t
- 2 = ln
5
100
s
et>5 - 2 =
100
s
et>5
=
100
e2
s = A 13.53et>5 B m
When s = 400 m,
400 = 13.53et>5
t = 16.93 s = 16.9 s
The s–t graph is shown in Fig. a.
a s 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 s = 100 m = 0.04(100) = 4 m>s2
a s = 400 m = 0.04(400) = 16 m>s2
The a–s graph is shown in Fig. b.
Ans.
400
12–71.
The a–t graph for a car is shown. Construct the v–t and s–t graphs if the car starts from
rest at t = 0. At what time t' does the car stop?
Given:
a1
5
m
2
s
2
a2
m
2
s
t1
10 s
Solution:
a1
k
t1
2
3
ap1 ( t)
kt
vp1 ( t)
t
k
2
ap2 ( t)
a2
vp2 ( t)
vp1 t1 a2 t t1
sp2 ( t)
sp1 t1 vp1 t1 t t1 Guess
W1
t'
12 s
Given
0 0.01t1 t1
W2
sp1 ( t)
vp2 ( t' )
0
t
k
6
t'
1
2
a2 t t1
2
Find ( t' )
t'
22.5 s
t1 1.01t1 t'
Velocity (m/s)
30
vp1 W 1 20
vp2 W 2
10
0
0
5
10
15
W 1 W 2
Time (s)
20
25
Ans.
Distance (m)
300
sp1 W 1 200
sp2 W 2
100
0
0
5
10
15
W1 W2
Time (s)
20
25
12–72.
The v-s graph for an airplane traveling on a straight runway is shown. Determine the
acceleration of the plane at s = s3 and s = s4. Draw the a-s graph.
Given:
s1
100 m
s4
150 m
s2
200 m
v1
40
s3
50 m
v2
m
s
m
50
s
Solution:
a
0 s3 s1
a3
s1 s4 s2
a4
v
dv
ds
§ s3 · § v1 ·
¨ ¸ v1¨ ¸
© s1 ¹ © s1 ¹
a3
s4 s1
ª
º v2 v1
v2 v1 »
«v1 s2 s1
¬
¼ s2 s1
a4
8.00
m
Ans.
2
s
4.50
m
Ans.
2
s
The graph
2
V1
0 0.01s1 s1
a1 V 1
V2
s1 1.01s1 s2
a2 V 2
V 1 v1 s2
s1 s1 m
V 2 s1
º v2 v1 s2
ª
v2 v1 »
«v1 s2 s1
¬
¼ s2 s1 m
Acceleration in m/s^2
20
15
a1 V 1
a2 V 2
10
5
0
0
50
100
V1 V2
Distance in m
150
200
12–73.
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.
SOLUTION
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
v = 2v2x + v2y = 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.)
12–74.
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.
SOLUTION
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
t
r
L0
dr =
L0
C 3i + (6 - 2t)j D dt
r = c 3ti + A 6t - t2 B j d m
When t = 1 s and 3 s,
r t = 1 s = 3(1)i + C 6(1) - 12 D j = [3i + 5j] m>s
r 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 t = 3 s - r t = 1 s
= (9i + 9j) - (3i + 5j)
= {6i + 4j} m
Ans.
12–75
§
bt
dt
2
·
A particle is traveling with a velocity of v © a t e i c e j¹. Determine the magnitude of
the particle’s displacement from t = 0 to t1. Use Simpson’s rule with n steps to evaluate the
integrals. What is the magnitude of the particle’s acceleration when t = t2.
Given:
a
m
3
3
s
n
0.2
b
1
s
c
4
m
s
d
1
0.8
t1
2
t2
3s
2s
s
2
100
Displacement
t1
t1
´
bt
µ a t e dt
¶0
x1
d1
2
x1
2
x1 y1
d1
7.34 m
y1
´
2
µ c ed t dt
¶0
y1
3.96 m
8.34 m
Ans.
Acceleration
ax
d
bt
a te
dt
ay
d § dt
©c e
dt
ax2
0.14
2·
¹
m
2
s
a
bt
e
2 t
dt
bt
ab te
ax2
e
t2
2
2
2c d t e
ay2
·
¨ b t2¸
2
©
¹
b t2§ 1
a
ay2
0.52
m
2
s
a2
2
d t2
2c d t2 e
2
ax2 ay2
a2
0.541
m
2
s
Ans.
12–76.
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?
SOLUTION
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
t
r
L0
dr =
r = c
L0
2
3
A 16t i + 4t j + (5t + 2)k B dt
5
16 3
t i + t4j + a t2 + 2tb k d m
3
2
When t = 2 s,
r =
16 3
5
A 2 B i + A 24 B j + c A 22 B + 2(2) d k = {42.7i + 16.0j + 14.0k} m.
3
2
Thus, the coordinate of the particle is
(42.7, 16.0, 14.0) m
Ans.
12–77.
The car travels from A to B, and then from B to C, as shown
in the figure. Determine the magnitude of the displacement of
the car and the distance traveled.
2 km
A
B
3 km
SOLUTION
Displacement:
¢r = {2i - 3j} km
¢r = 222 + 32 = 3.61 km
Ans.
Distance Traveled:
d = 2 + 3 = 5 km
C
Ans.
12–78.
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?
SOLUTION
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 = 3.606 km = 3.61 km
Ans.
Average Velocity and Speed: The total time is ¢t = 5 + 8 + 10 = 23 min = 1380 s
The magnitude of average velocity is
vavg
3.606 A 103 B
¢r
=
= 2.61 m>s
=
¢t
1380
Ans.
and the average speed is
A vsp B avg =
9 A 103 B
s
=
= 6.52 m>s
¢t
1380
Ans.
12–79.
A car traveling along the straight portions of the road has
the velocities indicated in the figure when it arrives at
points A, B, and C. If it takes 3 s to go from A to B, and then
5 s to go from B to C, determine the average acceleration
between points A and B and between points A and C.
y
vC
x
vB
B
SOLUTION
vA = 20 i
vB = 21.21 i + 21.21j
vA
A
vC = 40i
aAB =
21.21 i + 21.21 j - 20 i
¢v
=
¢t
3
aAB = { 0.404 i + 7.07 j } m>s2
aAC =
Ans.
40 i - 20 i
¢v
=
¢t
8
aAC = { 2.50 i } m>s2
Ans.
20 m/s
45
30 m/s
C
40 m/s
12–80.
A particle travels along the curve from A to B in 2 s. It takes
4 s for it to go from B to C and then 3 s to go from C to D.
Determine its average speed when it goes from A to D.
y
D
B
5m
15 m
C
SOLUTION
10 m
1
1
sT = (2p)(10)) + 15 + (2p(5)) = 38.56
4
4
vsP =
sT
38.56
= 4.28 m>s
=
tt
2 + 4 + 3
A
Ans.
x
12–81.
The position of a crate sliding down a ramp is given by
x = (0.25t3) m, y = (1.5t2) m, z = (6 - 0.75t5>2) m, where t
is in seconds. Determine the magnitude of the crate’s
velocity and acceleration when t = 2 s.
SOLUTION
Velocity: By taking the time derivative of x, y, and z, we obtain the x, y, and z
components of the crate’s velocity.
d
#
vx = x =
A 0.25t3 B = A 0.75t2 B m>s
dt
d
#
vy = y =
A 1.5t2 B = A 3t B m>s
dt
d
#
vz = z =
A 6 - 0.75t5>2 B =
dt
A - 1.875t3>2 B m>s
When t = 2 s,
vx = 0.75 A 22 B = 3 m>s
vz = - 1.875 A 2 B 3>2 = - 5.303 m/s
vy = 3(2) = 6 m>s
Thus, the magnitude of the crate’s velocity is
v = 2vx 2 + vy 2 + vz 2 = 232 + 62 + ( - 5.303)2 = 8.551 m>s = 8.55 m
Ans.
Acceleration: The x, y, and z components of the crate’s acceleration can be obtained
by taking the time derivative of the results of vx, vy, and vz, respectively.
d
#
ax = vx =
A 0.75t2 B = (1.5t) m>s2
dt
d
#
ay = vy =
(3t) = 3 m>s2
dt
d
#
az = vz =
A - 1.875t3>2 B =
dt
A - 2.815t1>2 B m>s2
When t = 2 s,
ax = 1.5(2) = 3 m>s2
ay = 3 m>s2
az = - 2.8125 A 21>2 B = - 3.977 m>s2
Thus, the magnitude of the crate’s acceleration is
a = 2ax 2 + ay 2 + az 2 = 232 + 32 + ( - 3.977)2 = 5.815 m>s2 = 5.82 m>s
Ans.
12–82.
A rocket is fired from rest at x = 0 and travels along a
parabolic trajectory described by y2 = [120(103)x] m. If the
1
x component of acceleration is ax = a t2 b m>s2, where t is
4
in seconds, determine the magnitude of the rocket’s velocity
and acceleration when t = 10 s.
SOLUTION
Position: The parameter equation of x can be determined by integrating ax twice
with respect to t.
L
dvx =
L
vx
t
dvx =
L0
vx = a
L
1 2
t dt
L0 4
1 3
t b m>s
12
dx =
x
L0
axdt
L
vxdt
t
dx =
x = a
1 3
t dt
L0 12
1 4
t bm
48
Substituting the result of x into the equation of the path,
y2 = 120 A 103 B a
1 4
t b
48
y = A 50t2 B m
Velocity:
d
#
vy = y =
A 50t2 B = A 100t B m>s
dt
When t = 10 s,
vx =
1
A 103 B = 83.33 m>s
12
vy = 100(10) = 1000 m>s
Thus, the magnitude of the rocket’s velocity is
v = 2vx 2 + vy2 = 283.332 + 10002 = 1003 m>s
Ans.
Acceleration:
#
d
ay = vy =
(100t) = 100 m>s2
dt
When t = 10 s,
ax =
1
A 102 B = 25 m>s2
4
Thus, the magnitude of the rocket’s acceleration is
a = 2ax 2 + ay2 = 2252 + 1002 = 103 m>s2
Ans.
12–83.
The flight path of the helicopter as it takes off from A is defined by the parametric equations
x = bt2 and y = ct3. Determine the distance the helicopter is from point A and the magnitudes
of its velocity and acceleration when t = t1.
Given:
b
2
m
2
s
t1
c
0.04
m
3
s
10 s
Solution:
r1
§ b t12 ·
¨
¸
¨ 3¸
© c t1 ¹
r1
§ 200 ·
¨
¸m
© 40 ¹
r1
204 m
v1
§ 2b t1 ·
¨
¸
¨ 3c t12 ¸
©
¹
a1
§ 2b ·
¨
¸
© 6c t1 ¹
v1
§ 40 · m
¨ ¸
© 12 ¹ s
a1
§ 4 ·m
¨ ¸ 2
© 2.4 ¹ s
v1
41.8
m
s
a1
4.66
m
2
s
Ans.
12–84.
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
y
π x)
c sin ( ––
L
x
c
L
SOLUTION
y = c sin a
p
xb
L
p
p
#
#
ca cos xb x
y =
L
L
vy =
p
p
c vx a cos x b
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
-2
v0 pc
p
p 2
p
vy =
acos xb B 1 + a cb cos2 a x b R
L
L
L
L
Ans.
c
L
12–85.
A particle travels along the curve 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.
y
B
20 m
SOLUTION
A
Time from B to C is 3 - 1 = 2 s
vang =
(rAC - rAB)
40i - (20i + 20j)
¢r
=
=
= {10i - 10j} m>s
¢t
¢t
2
Ans.
C
x
12–86.
y
When a rocket reaches an altitude of 40 m it begins to travel
along the parabolic path 1y - 4022 = 160x, where the
coordinates are measured in meters. If the component of
velocity in the vertical direction is constant at vy = 180 m>s,
determine the magnitudes of the rocket’s velocity and
acceleration when it reaches an altitude of 80 m.
(y
40)2
160x
SOLUTION
vy = 180 m>s
40 m
(y - 40)2 = 160 x
2(y - 40)vy = 160vx
(1)
2(80 - 40)(180) = 160vx
vx = 90 m>s
v = 2902 + 1802 = 201 m>s
ay =
d vy
dt
Ans.
= 0
From Eq. 1,
2 v2y + 2(y - 40)ay = 160 ax
2(180)2 + 0 = 160 ax
ax = 405 m>s2
a = 405 m>s2
Ans.
x
12–87.
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
D
C
SOLUTION
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),
23
1
(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 vx 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 y
2
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.
v
B
x2
4
y2
1
x
10 m/s
12–88.
Show that if a projectile is fired at an angle T from the horizontal with an initial velocity v0, the
maximum range the projectile can travel is given by R max = v02/g, where g is the acceleration of
gravity. What is the angle T for this condition?
After time t,
Solution:
v0 cos T t
x
v0 sin T t y
x
t
1
2
gt
v0 cos T
2
2
x tan T y
gx
2
2 v0 cos T
2
Set y = 0 to determine the range, x = R :
2
R
2
2 v0 sin T cos T
v0 sin 2 T
g
g
R max occurs when sin 2 T
1 or,
T
45deg
2
This gives:
R max
v0
g
Q.E.D
Ans.
12–89.
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
A
uA
1.2 m
50 m
SOLUTION
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
uA = 30.5°
vA = 23.2 m>s
Ans.
B
12–90.
Determine the minimum initial velocity v0 and the
corresponding angle u0 at which the ball must be kicked in
order for it to just cross over the 3-m high fence.
v0
3m
u0
6m
SOLUTION
Coordinate System: The x - y coordinate system will be set so that its origin
coincides with the ball’s initial position.
x-Motion: Here, (v0)x = v0 cos u, x0 = 0, and x = 6 m. Thus,
+ B
A:
x = x0 + (v0)xt
6 = 0 + (v0 cos u)t
t =
6
v0 cos u
(1)
y-Motion: Here, (v0)x = v0 sin u, ay = - g = - 9.81 m > s2, and y0 = 0. Thus,
A+cB
y = y0 + (v0)y t +
1
a t2
2 y
3 = 0 + v0 (sin u) t +
1
(-9.81)t2
2
3 = v0 (sin u) t - 4.905t2
(2)
Substituting Eq. (1) into Eq. (2) yields
v0 =
58.86
B sin 2u - cos2 u
(3)
From Eq. (3), we notice that v0 is minimum when f(u) = sin 2u - cos2 u is
df(u)
= 0
maximum. This requires
du
df(u)
= 2 cos 2u + sin 2u = 0
du
tan 2u = - 2
2u = 116.57°
u = 58.28° = 58.3°
Ans.
Substituting the result of u into Eq. (2), we have
(v0)min =
58.86
= 9.76 m>s
B sin 116.57° - cos2 58.28°
Ans.
12–9 1.
The pitching machine is adjusted so that the
baseball is launched with a speed of W" 30 ms. If the ball
strikes the ground at B, determine the two possible angles u"
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, (v")Y 30 cos u", Y" 0 and Y# 30 m. Thus,
Y# Y"
30 0
U (v")YU
30 cos u"U
1
cos u"
(1)
y-Motion: Here, (v")Z 30 sin u", BZ H 9.81 ms2, and Z# 1.2 m. Thus,
D
Z# Z"
1.2 0
1
B U2
2 Z
1
( 9.81)U2
30 sin u"U
2
(v")ZU
4.905U2 30 sin u" U 1.2 0
(2)
Substituting Eq. (1) into Eq. (2) yields
4.905 A
2
1
1
I 30 sin u" A
I 1.2 0
cos u"
cos u"
1.2 cos2 u"
30 sin u" cos u" 4.905 0
Solving by trial and error,
u" 7.19° and 80.5°
Ans.
12–92.
The girl always throws the toys at an angle of 30° from
point A as shown.Determine the time between throws so
that both toys strike the edges of the pool B and C at the
same instant.With what speed must she throw each toy?
A
30°
1m
B
2.5 m
SOLUTION
4m
To strike B:
+ )s = s + v t
(:
0
0
2.5 = 0 + vA cos 30° t
(+ c ) s = s0 + v0 t +
1 2
ac t
2
0.25 = 1 + vA siv 30° t -
1
(9.81)t2
2
Solving
t = 0.6687 s
(vA)B = 4.32 m>s
Ans.
To strike C:
+ )s = s + v t
(:
0
0
4 = 0 + vA cos 30° t
(+ c ) s = s0 + v0 t +
1 2
a t
2 c
0.25 = 1 + vA siv 30° t -
1
(9.81)t2
2
Solving
t = 0.790 s
(vA)C = 5.85 m>s
Ans.
Time between throws:
¢t = 0.790 s - 0.6687 s = 0.121 s
Ans.
C
0.25 m
12–93.
The balloon A is ascending at rate vA and is being carried horizontally by the wind at vw. If a
ballast bag is dropped from the balloon when the balloon is at height h, determine the time
needed for it to strike the ground. Assume that the bag was released from the balloon with the
same velocity as the balloon. Also, with what speed does the bag strike the ground?
Given:
vA
12
km
hr
vw
20
km
hr
h
50 m
g
9.81
m
2
s
Solution:
Thus
ax
0
ay
g
vx
vw
vy
g t vA
sx
vw t
sy
1 2
g t vA t h
2
0
vx
1 2
g t vA t h
2
vw
vy
g t vA
t
2
vA vA 2g h
g
v
2
2
vx vy
t
3.55 s
v
32.0
m
s
Ans.
Ans.
12–94.
The projectile is launched with a
velocity 2 v0 . Determine the range R,
the maximum height h attained, and
the time of flight. Express the results
in terms of the angle θ and v0. The
acceleration due to gravity is g.
2v0
Solution:
ax = 0
vx =
ay = − g
2v0 cos ( θ )
sx = 2v0 cos ( θ ) t
0 =
−1
2
g (tR) + 2v0 sin ( θ ) tR
2
R = 2v0 cos ( θ ) tR
h=
−1
2
vy = −g t + 2v0 sin ( θ )
sy =
−1
2
g t + 2v0 sin ( θ ) t
2
tR=
R=
h=
Ans.
g
8 v0
2
g
sin ( θ ) cos ( θ )
2v0 sin ( θ )
2
2
t
⎛tR⎞
g ⎜ ⎟ + 2v0 sin ( θ ) R
2
2
⎝ ⎠
4 v0 sin ( θ )
g
Ans.
2
Ans.
12–95.
A projectile is given a velocity v0 at an angle f above the
horizontal. Determine the distance d to where it strikes the
sloped ground. The acceleration due to gravity is g.
y
v0
φ
SOLUTION
+ B
A:
d cos u = 0 + v0 (cos f) t
s = s0 + v0 t +
1 2
a t
2 c
d sin u = 0 + v0 (sin f)t +
1
( - g) t2
2
Thus,
d sin u = v0 sin fa
d cos u
1
d cos u 2
b - ga
b
v0 cos f
2 v0 cos f
sin u = cos u tan f -
gd cos2 u
2v20 cos2 f
d = (cos u tan f - sin u)
d =
x
d
s = s0 + v0 t
A+cB
θ
2v20 cos2 f
g cos2 u
v 20
sin 2f - 2 tan u cos2 f
g cos u
Ans.
12–96.
A projectile is given a velocity v0 . Determine the angle f at
which it should be launched so that d is a maximum. The
acceleration due to gravity is g.
y
v0
φ
SOLUTION
+ B
A:
d cos u = 0 + v0 (cos f) t
sy = s0 + v0 t +
1 2
at
2 c
d sin u = 0 + v0 (sin f)t +
1
( - g)t2
2
Thus,
d sin u = v0 sin f a
d cos u
1
d cos u 2
b - ga
b
v0 cos f
2 v0 cos f
sin u = cos u tan f -
gd cos2 u
2v 20 cos2 f
d = (cos u tan f - sin u)
d =
2v 20 cos2 f
g cos2 u
v 20
A sin 2f - 2 tan u cos2 f B
g cos u
Require:
d(d)
v 20
=
C cos 2f(2) - 2 tan u(2 cos f)(- sin f) D = 0
df
g cos u
cos 2f + tan u sin 2f = 0
sin 2f
tan u + 1 = 0
cos 2f
tan 2f = - ctn u
f =
x
d
sx = s0 + v0 t
A+cB
θ
1
tan - 1 (- ctn u)
2
Ans.
12–97.
Measurements of a shot recorded on a videotape during a basketball game are shown. The ball
passed through the hoop even though it barely cleared the hands of the player B who attempted
to block it. Neglecting the size of the ball, determine the magnitude vA of its initial velocity and
the height h of the ball when it passes over player B.
Given:
a
2.1 m
b
7.5 m
c
1.5 m
d
3m
T
30 deg
g
9.81
m
2
s
Solution:
Guesses
Given
vA
bc
d
§ vA ·
¨ ¸
¨ tB ¸
¨ tC ¸
¨ ¸
©h¹
3
m
s
tB
1s
vA cos T tC
g 2
tC vA sin T tC a
2
Find vA tB tC h
§ tB ·
¨ ¸
© tC ¹
tC
h
1s
3.5 m
b
vA cos T tB
h
g 2
tB vA sin T tB a
2
§ 0.78 ·
¨
¸s
© 0.94 ¹
vA
11.1
m
s
h
3.45 m
Ans.
A projectile is fired from the platform at B. The
12–98.
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.
Y$ Y"
20 0
(v")YU
v" cos 30° U
(1)
y-Motion: Here, Z" 1.8, (v")Z v" sin 30°, and BZ H 9.81 ms2. Thus,
D
1
B U2
2 Z
Z$ Z"
(v")ZU
10 1.8
v" sin 30°(U)
1
( 9.81)(U)2
2
Thus,
10 1.8 A
20 sin 30°
I (U) 4.905(U)2
cos 30°(U)
U 0.8261 s
So that
v" 20
28.0 ms
cos 30°(0.8261)
10 m
30
1.8 m
20 m
x-Motion: Here, Y" 0 and Y$ 20 m. Thus,
A
Ans.
12–99.
The snowmobile is traveling at speed v0 when it leaves the embankment at A. Determine the
time of flight from A to B and the range R of the trajectory.
Given:
m
s
v0
10
T
40 deg
c
3
d
4
g
9.81
m
2
s
Solution:
R
Guesses
Given
§R·
¨ ¸
©t¹
R
1m
t
1s
§ c · R
¨ ¸
© d ¹
v0 cos T t
Find ( R t)
t
2.48 s
R
19.01 m
§ g · t2 v sin T t
¨ ¸
0
© 2 ¹
Ans.
12–100.
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.
vA
uA
A
4m
3
5
4
SOLUTION
+ B
A:
100 m
s = v0 t
4
100 a b = vA cos 25°tAB
5
A+cB
s = s0 + v0 t +
B
1
ac t2
2
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.
12–101.
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 speed at which he
strikes the ground.
vA
uA
A
4m
3
5
4
100 m
SOLUTION
Coordinate System: x - y coordinate system will be set with its origin to coincide
with point A as shown in Fig. a.
4
x-motion: Here, xA = 0, xB = 100 a b = 80 m and (vA)x = vA cos 25°.
5
+ B
A:
xB = xA + (vA)xt
80 = 0 + (vA cos 25°)t
80
t =
vA cos 25°
(1)
3
y-motion: Here, yA = 0, yB = - [4 + 100 a b ] = - 64 m and (vA)y = vA sin 25°
5
and ay = - g = - 9.81 m>s2.
A+ c B
yB = yA + (vA)y t +
1
a t2
2 y
- 64 = 0 + vA sin 25° t +
1
(- 9.81)t2
2
4.905t2 - vA sin 25° t = 64
(2)
Substitute Eq. (1) into (2) yieldS
4.905 ¢
¢
2
80
80
≤ = vA sin 25° ¢
≤ = 64
vA cos 25°
yA cos 25°
2
80
≤ = 20.65
vA cos 25°
80
= 4.545
vA cos 25°
vA = 19.42 m>s = 19.4 m>s
Substitute this result into Eq. (1),
t =
80
= 4.54465
19.42 cos 25°
Ans.
B
12–101. continued
Using this result,
A+ c B
(vB)y = (vA)y + ay t
= 19.42 sin 25° + ( -9.81)(4.5446)
= - 36.37 m>s = 36.37 m>s T
And
+ B
A:
(vB)x = (vA)x = vA cos 25° = 19.42 cos 25° = 17.60 m>s :
Thus,
vB = 2(vB)2x + (vB)2y
= 236.372 + 17.602
= 40.4 m>s
Ans.
12– 102. The path of a particle is defined by Z2 4LY, and
the component of velocity along the y axis is WZ DU, where
both k and c are constants. Determine the x and y
components of acceleration when Z Z0.
Z2 4LY
2ZvZ 4LvY
2v2Z
2ZBZ 4LBY
vZ DU
BZ D
2(DU)2
BY Ans.
2ZD 4LBY
D
Z
2L
DU2 Ans.
12–103.
The drinking fountain is designed such that the nozzle is located from the edge of the basin as
shown. Determine the maximum and minimum speed at which water can be ejected from the
nozzle so that it does not splash over the sides of the basin at B and C.
Given:
T
40 deg
a
50 mm
g
9.81
m
b
100 mm
c
250 mm
2
s
Solution:
Guesses
Given
vmin
1
m
s
tmin
1s
vmax
1
m
s
tmax
1s
b
vmin sin T tmin
bc
§ tmin ·
¨
¸
¨ tmax ¸
¨ vmin ¸
¨
¸
© vmax ¹
vmax sin T tmax
Find tmin tmax vmin vmax
a vmin cos T tmin 1
2
g tmin
2
a vmax cos T tmax 1
2
g tmax
2
§ tmin ·
¨
¸
© tmax ¹
§ 0.186 ·
¨
¸s
© 0.309 ¹
§ vmin ·
¨
¸
© vmax ¹
§ 0.838 · m
¨
¸
© 1.764 ¹ s
0
0
Ans.
12–104.
The projectile is launched with a velocity v0. Determine the
range R, the maximum height h attained, and the time of
flight. Express the results in terms of the angle u and v0. The
acceleration due to gravity is g.
y
v0
u
h
x
R
SOLUTION
+ B
A:
s = s0 + v0 t
R = 0 + (v0 cos u)t
A+cB
s = s0 + v0 t +
1
a t2
2 c
0 = 0 + (v0 sin u) t +
0 = v0 sin u -
R =
t =
=
A+cB
1
( - g)t2
2
1
R
(g) ¢
≤
2
v0 cos u
v20
sin 2u
g
Ans.
v20 (2 sin u cos u)
R
=
v0 cos u
v0 g cos u
2v0
sin u
g
Ans.
v2 = v20 + 2ac(s - s0)
0 = (v0 sin u)2 + 2(- g)(h - 0)
h =
v20
sin2 u
2g
Ans.
12–105.
The projectile is launched from a height h with
a velocity v0. Determine the range R.
Solution:
ax
0
ay
g
vx
v0 cos T
vy
g t v0 sin T
sx
v0 cos T t
sy
1 2
g t v0 sin T t h
2
When it hits
R
v0 cos T t
0
1 2
g t v0 sin T t h
2
t
R
v0 cos T
R
g §
¨
2 © v0 cos T
2
R
·
¸ v0 sin T v cos T h
0
¹
Solving for R we find
2
R
v0 cos T
g
2
2g h
§¨
2
tan T tan T 2
¨
v0 cos T
©
¸·
2¸
¹
Ans.
12–106.
The velocity of a particle is given by v = [at2i + bt3j + (ct + d)k]. If the particle is at the origin
when t = 0, determine the magnitude of the particle's acceleration when t = t1. Also, what is the
x, y, z coordinate position of the particle at this instant?
Given:
a
12
m
b
3
5
s
m
c
4
m
6
d
2
s
s
3
m
s
Solution:
Acceleration
ax
2a t1
ax
m
96.00
2
s
ay
2
3b t1
ay
240.00
m
2
s
az
c
az
6.00
m
2
s
2
2
ax ay az
amag
2
amag
258.6
m
2
Ans.
s
Postition
x
a 3
t1
3
x
256 .00 m
Ans.
y
b 4
t1
4
y
320 .00 m
Ans.
z
c 2
t1 d t1
2
z
60 .00 m
Ans.
t1
4s
12–107. The golf ball is hit at A with a speed of
W" 40 ms 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
d
A
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
5
x-Motion: Here, (v")Y 40 cos 30° 34.64 ms, Y" 0, and Y# E A
2
25
0.9806E. Thus,
Y# Y"
1
I
(v")YU
0.9806E 0
34.64U
U 0.02831E
(1)
y-Motion: Here, (v")Z 40 sin 30° 20 ms, Z" 0, Z# E A
2
0.1961E, and BZ H 9.81 ms .
1
A 252
1
I
Thus,
D
Z# Z"
(v")ZU
0.1961E 0
2
4.905U 20U
20U
1
B U2
2 Z
1
( 9.81)U2
2
0.1961E 0
(2)
Substituting Eq. (1) into Eq. (2) yields
4.905(0.02831E)2 20(0.02831E)
0.1961E 0
3.9303 10 3 E2 0.37002E 0
E 3.9303 10 3 E 0.37002 0
Since E ; 0, then
3.9303 10 3 E 0.37002 0
E 94.1 m
5
Ans.
1
12–108.
The man at A wishes to throw two darts at the target at B so
that they arrive at the same time. If each dart is thrown with
a speed of 10 m>s, determine the angles uC and uD at which
they should be thrown and the time between each throw.
Note that the first dart must be thrown at uC 1 7uD2, then
the second dart is thrown at uD .
5m
uC
A
s = s0 + v0t
5 = 0 + (10 cos u) t
(+ c )
(1)
v = v0 + act
-10 sin u = 10 sin u - 9.81 t
t =
2(10 sin u)
= 2.039 sin u
9.81
From Eq. (1),
5 = 20.39 sin u cos u
Since
sin 2u = 2 sin u cos u
sin 2u = 0.4905
The two roots are uD = 14.7°
Ans.
uC = 75.3°
Ans.
From Eq. (1): tD = 0.517 s
tC = 1.97 s
So that ¢t = tC - tD = 1.45 s
D
B
SOLUTION
+ )
(:
uD
C
Ans.
12–109.
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
u1
u2
y
SOLUTION
Vertical Motion: For the first ball, the vertical component of initial velocity is
(v0)y = v0 sin u1 and the initial and final vertical positions are (s0)y = 0 and sy = y,
respectively.
(+ c)
sy = (s0)y + (v0)y t +
1
(a ) t2
2 cy
y = 0 + v0 sin u1t1 +
1
( -g)t21
2
(1)
For the second ball, the vertical component of initial velocity is (v0)y = v0 sin u2 and
the initial and final vertical positions are (s0)y = 0 and sy = y, respectively.
(+ c)
sy = (s0)y + (v0)y t +
1
(a ) t2
2 cy
y = 0 + v0 sin u2t2 +
1
( -g)t22
2
(2)
Horizontal Motion: For the first ball, the horizontal component of initial velocity is
(v0)x = v0 cos u1 and the initial and final horizontal positions are (s0)x = 0 and
sx = x, respectively.
+ B
A:
sx = (s0)x + (v0)x t
x = 0 + v0 cos u1 t1
(3)
For the second ball, the horizontal component of initial velocity is (v0)x = v0 cos u2
and the initial and final horizontal positions are (s0)x = 0 and sx = x, respectively.
+ B
A:
sx = (s0)x + (v0)x t
x = 0 + v0 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
v0 t1 sin u1 - v0 t2 sin u2 =
1
g A t21 - t22 B
2
(6)
Solving Eq. [5] into [6] yields
t1 =
t2 =
2v0 cos u2 sin(u1 - u2)
g(cos2 u2 - cos2 u1)
2v0 cos u1 sin(u1 - u2)
g(cos2 u2 - cos2u1)
Thus, the time between the throws is
¢t = t1 - t2 =
=
2v0 sin(u1 - u2)(cos u2 - cos u1)
g(cos2 u2 - cos2 u1)
2v0 sin (u1 - u2)
g(cos u2 + cos u1)
Ans.
x
12–110.
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
B
SOLUTION
Vertical Motion: The vertical component of initial velocity is (v0)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 + (v0)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 (v0)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 + (v0)x t
R = 0 + 1.732(0.6867) = 1.19 m
Ans.
If sx = R + 1, then
+ B
A:
sx = (s0)x + (v0)x t
R + 1 = 0 + 1.732(0.6867)
R = 0.189 m
Thus, Rmin = 0.189 m, Rmax = 1.19 m
Ans.
Ans.
R
1m
12–111. A projectile is fired with a speed of W 60 ms 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, vY 60 cos 60° 30 ms, Y0 0,
and U U1. Thus,
Y Y0
vYU
Y 0
30U1
(1)
For the motion of the second projectile, vY 60 cos u, Y0 0, and U U1 0.5.
Thus,
Y Y0
Y 0
vYU
60 cos u(U1 0.5)
(2)
y-Motion: For the motion of the first projectile, vZ 60 sin 60° 51.96 ms, Z0 0,
and BZ H 9.81 ms2. Thus,
D
Z Z0
Z 0
1 2
BU
2 Z
1
( 9.81)U1 2
51.96U1
2
vZU
Z 51.96U1 4.905U1 2
For the motion of the
BZ H 9.81 ms2. Thus,
D
Z Z0
Z 0
vZU
(3)
second
projectile,
vZ 60 sin u,
Z0 0,
and
1 2
BZU
2
1
( 9.81)(U1 0.5)2
2
60 sin u(U1 0.5)
Z (60 sin u)U1 30 sin u 4.905 U1 2
4.905U1 1.22625
(4)
Equating Eqs. (1) and (2),
30U1 60 cos u(U1 0.5)
U1 cos u
2 cos u 1
(5)
Equating Eqs. (3) and (4),
51.96U1 4.905U1 2 (60 sin u)U1 30 sin u 4.905U1 2
(60 sin u 47.06)U1 30 sin u
U1 30 sin u 1.22625
60 sin u 47.06
4.905U1 1.22625
1.22625
(6)
y
Equating Eqs. (5) and (6) yields
cos u
30 sin u 1.22625
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)),
U1 cos 57.57°
7.3998 s
2 cos 57.57° 1
Substituting this result into Eqs. (1) and (3),
Y 30(7.3998) 222 m
Ans.
Z 51.96(7.3998) 4.905 7.39982 116 m
Ans.
The boy at A attempts to throw a ball over the
12–112.
roof of a barn with an initial speed of V! 15 ms.
Determine the angle .! 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
vA
A
8m
.A
1m
d
Vertical Motion: The vertical component, of initial and final velocity are
(20)Y (15 sin .!) ms and 2Y 0, respectively. The initial vertical position is
(S0)Y 1 m.
C
2Y (20)
0 15 sin .!
C
SY (S0)Y
8 1
AC T
( 9.81)T
(20)Y T
15 sin .!T
1
(A ) T2
2 CY
1
( 9.81)T2
2
[1]
[2]
Solving Eqs. [1] and [2] yields
.! 51.38° 51.4°
Ans.
T 1.195 s
Horizontal Motion: The horizontal component of velocity is (20)X 2! cos .!
15 cos 51.38° 9.363 ms. The initial and final horizontal positions are (S0)X 0
and SX (D 4) m, respectively.
SX (S0)X
D
4 0
(20)X T
9.363(1.195)
D 7.18 m
Ans.
4m
12–113. The boy at A attempts to throw a ball over the
roof of a barn such that it is launched at an angle .! 40°.
Determine the minimum speed V! 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
vA
A
8m
.A
1m
d
Vertical Motion: The vertical components of initial and final velocity are
(20)Y (2! sin 40°) ms and 2Y 0, respectively. The initial vertical position is
(S0)Y 1 m.
C
2Y (20)
AC T
0 2! sin 40°
C
SY (S0)Y
8 1
( 9.81) T
(20)Y T
2! sin 40°T
1
(A ) T2
2 CY
1
( 9.81) T2
2
[1]
[2]
Solving Eqs. [1] and [2] yields
2! 18.23 ms 18.2 ms
T 1.195 s
Ans.
Horizontal Motion: The horizontal component of velocity is (20)X 2! cos .!
18.23 cos 40° 13.97 ms. The initial and final horizontal positions are (S0)X 0
and SX (D 4) m, respectively.
SX (S0)X
D
4 0
(20)X T
13.97(1.195)
D 12.7 m
Ans.
4m
12–114.
A car is traveling along a circular curve that has a radius of
50 m. If its speed is 16 m>s and is increasing uniformly at
8 m>s2, determine the magnitude of its acceleration at this
instant.
SOLUTION
v = 16 m>s
at = 8 m>s2
r = 50 m
an =
(16)2
v2
=
= 5.12 m>s2
r
50
a = 2(8)2 + (5.12)2 = 9.50 m>s2
Ans.
12–115.
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.
SOLUTION
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
v = 38.7 m>s
Ans.
12–116.
The car travels around the circular track having a radius r such that when it is at point A it has
a velocity v1 which is increasing at the rate dv/dt = kt. Determine the magnitudes of its
velocity and acceleration when it has traveled one-third the way around the track.
Given:
k
m
0.06
3
s
r
300 m
v1
5
m
s
Solution:
at ( t )
kt
v ( t)
v1 sp ( t)
k 2
t
2
v1 t k 3
t
6
t1
1s
Guess
sp t1
Given
2S r
3
t1
Find t1
t1
35.58 s
2
v1
v t1
v1
43.0
at1
m
s
at t1
a1
an1
6.52
m
2
s
v1
r
Ans.
a1
2
2
at1 an1
12–117.
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.
SOLUTION
at = ¢
2
1h
2000 km 1000 m
ba
b = 0.1543 m>s2
ba
2
1 km
3600 s
h
v = a
an =
1h
60 km 1000 m
ba
ba
b = 16.67 m>s
h
1 km
3600 s
v2
16.672
= 0.4630 m>s2
=
r
600
a = 2a2t + a2n = 20.15432 + 0.46302 = 0.488 m>s2
Ans.
12–118.
The truck travels in a circular path having a radius of 50 m at
a speed of v = 4 m>s. For a short distance from s = 0, its
#
speed is increased by v = 10.05s2 m>s2, where s is in meters.
Determine its speed and the magnitude of its acceleration
when it has moved s = 10 m.
.
v
v
50 m
SOLUTION
v dv = at ds
v
L4
10
v dv =
L0
0.5v 2 - 8 =
0.05s ds
0.05
(10)2
2
v = 4.583 = 4.58 m>s
an =
Ans.
v 2 (4.583)2
= 0.420 m>s2
=
r
50
at = 0.05(10) = 0.5 m>s2
a = 2(0.420)2 + (0.5)2 = 0.653 m>s2
(0.05s) m/s2
4 m/s
Ans.
12–119.
Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is
posted at v, determine the maximum acceleration experienced by the passengers.
Given:
km
hr
v
60
a
60 m
b
40 m
Solution:
Maximum acceleration
occurs where the radius of
curvature is the smallest.
In this case that happens
when y = 0.
x ( y)
U ( y)
§ y·
a 1¨ ¸
© b¹
2
1 x' ( y)
x'' ( y)
2
x' ( y)
d
x ( y)
dy
x'' ( y)
d
x' ( y)
dy
U min
U ( 0m)
U min
26.667 m
amax
10.42
3
2
amax
v
U min
m
2
s
Ans.
12–120.
Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is
posted at v, determine the minimum acceleration experienced by the passengers.
Given:
km
hr
v
60
a
60 m
b
40 m
Solution:
Minimum acceleration
occurs where the radius of
curvature is the largest. In
this case that happens
when x = 0.
y ( x)
U ( x)
§x·
b 1¨ ¸
© a¹
2
1 y' ( x)
y'' ( x)
2
y' ( x)
d
y ( x)
dx
y'' ( x)
d
y' ( x)
dx
U max
U ( 0m)
U max
90 m
amin
3.09
3
2
amin
v
U max
m
2
s
Ans.
12–121.
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
SOLUTION
B
Radius of Curvature:
x
1 2
y =
x
100
dy
1
=
x
dx
50
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
The angle that the tangent at B makes with the x axis is f = tan-1 ¢
Ans.
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 = 69.16°. Thus, the angle u that the roller coaster’s acceleration makes
b = tan-1 a
at
3
with the x axis is
u = a - f = 38.2° b
Ans.
12–122.
If the roller coaster starts from rest at A and its speed
increases at at = (6 – 0.06s) m>s2, determine the magnitude
of its acceleration when it reaches B where sB = 40 m.
y
y
1 x2
100
A
s
SOLUTION
B
Velocity: Using the initial condition v = 0 at s = 0,
x
v dv = at ds
v
L0
s
vdv =
L0
A 6 - 0.06s B ds
30 m
v = a 212s - 0.06s2 b m>s
(1)
Thus,
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
=
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 = 6 - 0.06(40) = 3.600 m>s2
an =
19.602
v2
= 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.
12–123.
The speedboat travels at a constant speed of 15 m> s while
making a turn on a circular curve from A to B. If it takes
45 s to make the turn, determine the magnitude of the
boat’s acceleration during the turn.
A
r
SOLUTION
Acceleration: During the turn, the boat travels s = vt = 15(45) = 675 m. Thus, the
675
s
radius of the circular path is r =
=
m. Since the boat has a constant speed,
p
p
at = 0. Thus,
a = an =
v2
=
r
152
= 1.05 m>s2
675
a
b
p
B
Ans.
12–124.
The car travels along the circular path such that its speed is
increased by a t = (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.
s
SOLUTION
v
L0
t
dv =
L0
0.5e t dt
ρ
v = 0.5(e t - 1)
t
18
L0
ds = 0.5
L0
(e t - 1)dt
18 = 0.5(e t - t - 1)
Solving,
t = 3.7064 s
v = 0.5(e 3.7064 - 1) = 19.85 m>s = 19.9 m>s
#
at = v = 0.5e t ƒ t = 3.7064 s = 20.35 m>s2
an =
Ans.
19.852
v2
= 13.14 m>s2
=
r
30
a = 2a2t + a2n = 220.352 + 13.142 = 24.2 m>s2
Ans.
30 m
18 m
12–125.
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
B
16 m
SOLUTION
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
dx2
= - 3.2 A 10-3 B
B1 + a
r =
2
dy 2
b
dx
d2y
dx2
B
d2y
3>2
2 3>2
c 1 + a -3.2 A 10-3 B x b d
=
2
2 - 3.2 A 10-3 B 2
16
4
= 324.58 m
x = 50 m
Acceleration:
vB 2
17.282
= 0.9194 m>s2
=
r
324.58
dv
= A 25 - 0.15s B A - 0.15 B = A 0.225s - 3.75 B m>s2
at = v
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.
1 2
x
625
s
A x
12–126.
y
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
B
16 m
SOLUTION
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
B1 + a
r =
2
dy 2 3>2
b R
dx
d2y
dx2
2 3>2
c1 + a -3.2 A 10-3 B x b d
=
2
- 3.2 A 10-3 B
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
16
Ans.
1 2
x
625
s
A x
12–127.
A train is traveling with a constant speed of 14 m/s along
the curved path. Determine the magnitude of the
acceleration of the front of the train, B, at the instant it
reaches point A 1y = 02.
y (m)
10 m
( —y )
x = 10 e 15
x (m)
SOLUTION
A
vt = 14 m/s
y
x = 10e(15)
y = 15 lna
x
b
10
dy
10
15
1
= 15 a b a b =
x
x
dx
10
d2y
dx2
= -
15
x2
At x = 10,
3
B1 + a
r =
2
at =
dy 2 2
b R
dx
d2y
dx2
2
C 1 + (1.5)2 D 2
3
=
|- 0.15|
= 39.06 m
dv
= 0
dt
an = a =
(14)2
v2
=
= 5.02 m s2
r
39.06
Ans.
B
12–128.
The car travels along the curve having a radius of R. If its speed is uniformly increased from
v1 to v2 in time t, determine the magnitude of its acceleration at the instant its speed is v3.
Given:
v1
15
m
s
v2
27
m
s
v3
20
m
s
t
R
3s
300 m
Solution:
at
v2 v1
t
2
an
v3
R
a
2
2
at an
a
4.22
m
2
s
Ans.
12–129. When the car reaches point A it has a speed of
25 ms. If the brakes are applied, its speed is reduced by
12
BU ( 14 U ) ms2. 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 ms at U 0 s,
Ev BU EU
U
v
(25 ms
Ev (0
1
U12 EU
4
1 32
U 4 ms
6
v 325 (1)
Position: Using the initial conditions T 0 when U 0 s,
(
ET (
T
(0
vEU
U
1
3 25 U 32 4 EU
6
(0
ET T 3 25U 1 52
U 4m
15
Acceleration: When the car reaches C, T$ 200
330.90 25U p
250 3 4 330.90 m. Thus,
6
1 52
U
15
Solving by trial and error,
U 15.942 s
Thus, using Eq. (1).
v$ 25 1
(15.942)32 14.391 ms
6
1
(B U)$ v 15.94212 0.9982 ms2
4
BO $ v$ 2
14.3912
0.8284 ms2
r
250
The magnitude of the car’s acceleration at C is
B 4 BU $ 2
BO $ 2 4( 0.9982)2
0.82842 1.30 ms2
Ans.
A
200 m
12–130. When the car reaches point A, it has a speed of
25 ms. If the brakes are applied, its speed is reduced by
AT (0.001S 1) ms2. Determine the magnitude of
acceleration of the car just before it reaches point C.
+ 250 m
C
B
Velocity: Using the initial condition v 25 ms at T 0 s,
30°
Dv ADS
S
v
'25 ms
vDv '0
v 0.001S2
(0.001S
2S
1)DS
625
Acceleration: When the car is at point C, S# 200
the speed of the car at C is
0.001 330.902 v# (AT)# V [0.001(330.90)
2
(AN)# 2(330.90)
1] )
2502 3 330.90 m. Thus,
6
625 8.526 ms2
0.6691 ms2
2
v#
8.526
0.2908 ms2
+
250
The magnitude of the car’s acceleration at C is
A (AT)# 2
(AN)# 2 ( 0.6691)2
0.29082 0.730 ms2
Ans.
A
200 m
12–131.
At a given instant the train engine at E has a speed of
20 m>s and an acceleration of 14 m>s2 acting in the
direction shown. Determine the rate of increase in the
train’s speed and the radius of curvature r of the path.
v
75
a
r
SOLUTION
at = 14 cos 75° = 3.62 m>s2
Ans.
an = 14 sin 75°
an =
(20)2
r
r = 29.6 m
14 m/s2
Ans.
E
20 m/s
12–132.
v
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
SOLUTION
Velocity: The speed v in terms of time t can be obtained by applying a =
5m
dv
.
dt
A
dv = adt
v
L0
t
dv =
L0
0.5et dt
v = 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
v =
ds
.
dt
ds = vdt
t
2.618 m
ds =
L0
L0
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
v = 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
v2
=
= 2.708 m>s2
r
5
The magnitude of the acceleration is
a = 2a2t + a2n = 24.1802 + 2.7082 = 4.98 m>s2
Ans.
u
12–133.
v
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
SOLUTION
5m
dv
Velocity: The speed v in terms of time t can be obtained by applying a =
.
dt
A
dv = adt
v
L0
t
dv =
L0
t
0.5e dt
v = 0.5 A et - 1 B
When t = 2 s,
v = 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 =
v2
3.1952
= 2.041 m>s2
=
r
5
The magnitude of the acceleration is
a = 2a2t + a2n = 23.6952 + 2.0412 = 4.22 m>s2
Ans.
u
12 –134. At a given instant, a car travels along a circular
curved road with a speed of 20 ms while decreasing its speed
at the rate of 3 ms2. If the magnitude of the car’s acceleration
is 5 ms2, determine the radius of curvature of the road.
Acceleration: Here, the
BU 3 ms2. Thus,
car’s
tangential
B 2BU 2
component
of
acceleration
of
BO 2
5 4 3 2
BO 2
BO 4 ms2
BO v2
r
4 202
r
r 100 m
Ans.
12–135.
A boat is traveling along a circular path having a radius of
20 m. Determine the magnitude of the boat’s acceleration
when the speed is v = 5 m>s and the rate of increase in the
#
speed is v = 2 m>s2.
SOLUTION
at = 2 m>s2
an =
y2
52
=
= 1.25 m>s2
r
20
a = 2a2t + a2n = 222 + 1.252 = 2.36 m>s2
Ans.
12–136.
Starting from rest, a bicyclist travels around a horizontal circular
path, r = 10 m, at a speed of v = 10.09t2 + 0.1t2 m>s,
where t is in seconds. Determine the magnitudes of his velocity
and acceleration when he has traveled s = 3 m.
SOLUTION
s
L0
t
ds =
L0
10.09t2 + 0.1t2dt
s = 0.03t3 + 0.05t2
When s = 3 m,
3 = 0.03t3 + 0.05t2
Solving,
t = 4.147 s
v =
ds
= 0.09t2 + 0.1t
dt
v = 0.09(4.147)2 + 0.1(4.147) = 1.96 m>s
at =
dv
= 0.18t + 0.1 `
= 0.8465 m>s2
dt
t = 4.147 s
an =
1.962
v2
= 0.3852 m>s2
=
r
10
a = 2a2t + a2n = 2(0.8465)2 + (0.3852)2 = 0.930 m>s2
Ans.
Ans.
12–137.
A particle travels around a circular path having a radius of
50 m. If it is initially traveling with a speed of 10 m> s and its
#
speed then increases at a rate of v = 10.05 v2 m>s2,
determine the magnitude of the particle’s acceleraton four
seconds later.
SOLUTION
Velocity: Using the initial condition v = 10 m>s at t = 0 s,
dt =
dv
a
t
L0
v
dt =
t = 20 ln
L10 m>s
dv
0.05v
v
10
v = (10et>20) m>s
When t = 4 s,
v = 10e4>20 = 12.214 m>s
Acceleration: When v = 12.214 m>s (t = 4 s),
at = 0.05(12.214) = 0.6107 m>s2
an =
(12.214)2
v2
=
= 2.984 m>s2
r
50
Thus, the magnitude of the particle’s acceleration is
a = 2at2 + an2 = 20.61072 + 2.9842 = 3.05 m>s2
Ans.
12–138.
When the bicycle passes point A, it has a speed of 6 m> s,
#
which is increasing at the rate of v = 10.52 m>s2. Determine
the magnitude of its acceleration when it is at point A.
y
y ⫽ 12 ln (
x
)
20
A
x
50 m
SOLUTION
Radius of Curvature:
y = 12 ln a
x
b
20
dy
1
1
12
= 12 a
ba b =
x
dx
x>20 20
d2y
dx2
= -
12
x2
B1 + a
r =
`
dy 2 3>2
b R
dx
d2y
`
dx2
B1 + ¢
=
`-
12 2 3>2
≤ R
x
12
`
x2
4
= 226.59 m
x = 50 m
Acceleration:
at = v = 0.5 m>s2
an =
v2
62
=
= 0.1589 m>s2
r
226.59
The magnitude of the bicycle’s acceleration at A is
a = 2at2 + an2 = 20.52 + 0.15892 = 0.525 m>s2
Ans.
12–139.
The motorcycle is traveling at a constant speed of 60 km> h.
Determine the magnitude of its acceleration when it is at
point A.
y
y2 ⫽ 2x
A
x
25 m
SOLUTION
Radius of Curvature:
y = 22x1>2
dy
1
= 22x - 1>2
dx
2
d2y
dx2
= -
1
22x - 3>2
4
B1 + a
r =
`
dy 2 3>2
b R
dx
d2y
`
dx2
1
2
2
B 1 + ¢ 22x - 1>2 ≤ R
=
3>2
1
` - 22x - 3>2 `
4
4
= 364.21 m
x = 25 m
Acceleration: The speed of the motorcycle at a is
v = ¢ 60
an =
km 1000 m
1h
≤¢
≤¢
≤ = 16.67 m>s
h
1 km
3600 s
v2
16.672
=
= 0.7627 m>s2
r
364.21
Since the motorcycle travels with a constant speed, at = 0. Thus, the magnitude of
the motorcycle’s acceleration at A is
a = 2at 2 + an2 = 202 + 0.76272 = 0.763 m>s2
Ans.
12–140.
The jet plane travels along the vertical parabolic path.
When it is at point A it has a speed of 200 m>s, which is
increasing at the rate of 0.8 m>s2. Determine the magnitude
of acceleration of the plane when it is at point A.
y
y ⫽ 0.4x2
A
SOLUTION
y = 0.4x2
10 km
dy
= 0.8x `
= 4
dx
x = 5 km
d2y
dx2
r =
= 0.8
x
5 km
[1 + (4)2]3/2
= 87.62 km
0.8
at = 0.8 m/s2
an =
(0.200)2
= 0.457(10 - 3) km>s2
87.62
an = 0.457 km/s2
a = 210.822 + 10.45722 = 0.921 m>s2
Ans.
12–141.
The ball is ejected horizontally from the tube with a speed
of 8 m>s. Find the equation of the path, y = f1x2, and then
find the ball’s velocity and the normal and tangential
components of acceleration when t = 0.25 s.
y
vA
A
SOLUTION
vx = 8 m>s
+ )
(:
s = v0t
x = 8t
A+cB
s = s0 + v0 t +
y = 0 + 0 +
1 2
a t
2 c
1
( - 9.81)t2
2
y = -4.905t2
x 2
y = -4.905 a b
8
y = -0.0766x2
(Parabola)
Ans.
v = v0 + act
vy = 0 - 9.81t
When t = 0.25 s,
vy = - 2.4525 m>s
v = 31822 + 12.452522 = 8.37 m>s
u = tan - 1 a
ax = 0
Ans.
2.4525
b = 17.04°
8
ay = 9.81 m>s2
an = 9.81 cos 17.04° = 9.38 m>s2
Ans.
at = 9.81 sin 17.04° = 2.88 m>s2
Ans.
8 m/s
x
12–142.
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
#
vA = 3 m>s2.
y
y = 0.01x2
A
36 m
SOLUTION
x
60 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
|d2y>dx2|
=
[1 + (0.02x)2]3>2
2
= 190.57 m
|0.02|
x = 60 m
To determine the normal acceleration, apply Eq. 12–20.
an =
v2
102
= 0.5247 m>s2
=
r
190.57
#
Here, at = vA = 3 m>s. Thus, the magnitude of acceleration is
a =
a2t + a2n =
32 + 0.52472 = 3.05 m s2
Ans.
12–143.
A particle P moves along the curve y = 1x2 - 42 m with a
constant speed of 5 m>s. Determine the point on the curve
where the maximum magnitude of acceleration occurs and
compute its value.
SOLUTION
y = (x2 - 4)
at =
dv
= 0,
dt
To obtain maximum a = an, r must be a minimum.
This occurs at:
x = 0,
y = -4 m
Ans.
Hence,
dy
`
= 2x = 0;
dx x = 0
d2y
dx2
= 2
3
c1 + a
rmin =
`
dy 2 2
b d
dx
d2y
dx2
`
(a)max = (an)max =
3
[1 + 0]2
1
=
=
020
2
v2
52
= 1 = 50 m>s2
rmin
2
Ans.
12–144.
The satellite S travels around the earth in a circular path with a constant speed v1. If the
acceleration is a, determine the altitude h. Assume the earth’s diameter to be d.
Units Used:
Mm
3
10 km
Given:
v1
20
a
2.5
Mm
hr
m
2
s
d
12713 km
Solution:
Guess
h
1 Mm
2
Given
a
v1
h
d
2
h
Find ( h)
h
5.99 Mm
Ans.
12–145.
The particle travels with a constant speed v along the curve. Determine the particle’s
acceleration when it is located at point x = x1.
Given:
mm
s
v
300
k
20 u 10 mm
3
x1
2
200 mm
Solution:
y ( x)
k
x
y' ( x)
d
y ( x)
dx
y'' ( x)
d
y' ( x)
dx
1 y' ( x)
y'' ( x)
U ( x)
2
§ sin T 1 ·
¨
¸
U x1 © cos T 1 ¹
3
T ( x)
atan ( y' ( x) )
2
a
v
a
§ 144 · mm
¨
¸
© 288 ¹ s2
T1
T x1
a
T1
322
mm
2
s
26.6 deg
Ans.
12–146.
The race car has an initial speed vA = 15 m>s at A. If it
increases its speed along the circular track at the rate
a t = 10.4s2 m>s2, where s is in meters, determine the time
needed for the car to travel 20 m. Take r = 150 m.
r
SOLUTION
s
A
n dn
at = 0.4s =
ds
a ds = n dn
n
s
0.4s ds =
L0
L15
n dn
n2 n
0.4s2 s
` =
`
2 0
2 15
n2
225
0.4s2
=
2
2
2
n2 = 0.4s2 + 225
n =
s
ds
= 20.4s2 + 225
dt
t
ds
L0 20.4s2 + 225
s
ds
L0 2s2 + 562.5
=
L0
dt
= 0.632 456t
1n (s + 2s2 + 562.5) `
s
= 0.632 456t
0
1n (s + 2s2 + 562.5) - 3.166 196 = 0.632 456t
At s = 20 m,
t = 1.21 s
Ans.
12–147.
A particle moves along the curve y = bsin(cx) with a constant speed v. Determine the normal and
tangential components of its velocity and acceleration at any instant.
Given:
v
2
m
s
b
1 m
1
c
m
Solution:
y
b sin ( c x)
y'
b c cos ( c x)
2
b c sin ( c x)
y''
3
U
1 y'
2
3
ª¬1 ( b c cos ( c x) ) 2º¼
y''
2
2
b c sin ( c x)
2
an
v b c sin ( c x)
3
ª¬1 ( b c cos ( c x) ) 2º¼
2
at
0
vt
0
vn
0
Ans.
12–148.
A particle travels along the path y = a + bx + cx 2, 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.
SOLUTION
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 =
vy =
v0
21 + b2
v0 b
21 + b2
v20
an =
r
Ans.
Ans.
dy 2 2
C 1 + A dx
B D
3
r =
`
d2y
dx
2
`
dy
= b + 2c x
dx
d2y
dx2
= 2c
At x = 0,
an =
r =
2 c v20
(1 + b2)3>2
(1 + b2)3>2
2c
Ans.
12–149.
The two particles A and B start at the origin O and travel in
opposite directions along the circular path at constant
speeds vA = 0.7 m>s and vB = 1.5 m>s, respectively.
Determine in t = 2 s, (a) the displacement along the path
of each particle, (b) the position vector to each particle, and
(c) the shortest distance between the particles.
y
SOLUTION
(a)
sA = 0.7(2) = 1.40 m
Ans.
sB = 1.5(2) = 3 m
Ans.
5m
B
(b)
1.40
uA =
= 0.280 rad. = 16.04°
5
uB =
A
vB
For A
x = 5 sin 16.04° = 1.382 = 1.38 m
y = 5(1 - cos 16.04°) = 0.1947 = 0.195 m
Ans.
For B
x = -5 sin 34.38° = -2.823 = -2.82 m
y = 5(1 -cos 34.38°) = 0.8734 = 0.873 m
rB = { - 2.82i + 0.873j} m
(c)
Ans.
¢r = rB - rA = {-4.20i + 0.678j} m
¢r = 31 - 4.2022 + 10.67822 = 4.26 m
x
O
vA
3
= 0.600 rad. = 34.38°
5
rA = {1.38i + 0.195j} m
1.5 m/s
Ans.
0.7 m/s
12–150.
The two particles A and B start at the origin O and travel in
opposite directions along the circular path at constant
speeds vA = 0.7 m>s and vB = 1.5 m>s, respectively.
Determine the time when they collide and the magnitude of
the acceleration of B just before this happens.
y
SOLUTION
st = 2p(5) = 31.4159 m
5m
sA = 0.7 t
B
sB = 1.5 t
A
vB
Require
0.7 t + 1.5 t = 31.4159
aB =
(1.5)2
v2B
= 0.45 m>s2
=
r
5
x
O
vA
sA + sB = 31.4159
t = 14.28 s = 14.3 s
1.5 m/s
Ans.
Ans.
0.7 m/s
12–151.
The position of a particle traveling along a curved path is
s = (3t3 - 4t2 + 4) m, where t is in seconds. When t = 2 s,
the particle is at a position on the path where the radius of
curvature is 25 m. Determine the magnitude of the
particle’s acceleration at this instant.
SOLUTION
Velocity:
v =
d 3
A 3t - 4t2 + 4 B = A 9t2 - 8t B m>s
dt
When t = 2 s,
v ƒ t = 2 s = 9 A 22 B - 8122 = 20 m>s
Acceleration:
at =
dv
d 2
=
A 9t - 8t B = A 18t - 8 B m>s2
ds
dt
at ƒ t = 2 s = 18(2) - 8 = 28 m>s2
an =
Av ƒ t = 2 sB2
r
=
202
= 16 m>s2
25
Thus,
a = 2at 2 + an2 = 2282 + 162 = 32.2 m>s2
Ans.
12–
Starting from rest the motorboat travels around
the circular path, + 50 m, at a speed V (0.8T) ms,
where t is in seconds. Determine the magnitudes of the
boat’s velocity and acceleration when it has traveled 20 m.
+
50 m
Velocity: The time for which the boat to travel 20 m must be determined first.
v
DS 2DT
T
20 m
'0
DS 0.8 TDT
'0
T 7.071 s
The magnitude of the boat’s velocity is
2 0.8 (7.071) 5.657 ms 5.66 ms
Ans.
Acceleration: The tangential accelerations is
AT 2 0.8 ms2
To determine the normal acceleration, apply Eq. 12–20.
AN 5.6572
22
0.640 ms2
+
50
Thus, the magnitude of acceleration is
A A2T
A2N 0.82
0.6402 1.02 ms2
Ans.
12–153. Starting from rest, the motorboat travels around
the circular path, + 50 m, at a speed V (0.2T2) ms,
where t is in seconds. Determine the magnitudes of the
boat’s velocity and acceleration at the instant T 3 s.
+ 50 m
Velocity: When T 3 s, the boat travels at a speed of
2 0.2 32 1. 80 ms
Ans.
Acceleration: The tangential acceleration is AT 2 (0.4T) ms2. When T 3 s,
AT 0.4 (3) 1.20 ms2
To determine the normal acceleration, apply Eq. 12–20.
AN 1.802
22
0.0648 ms2
+
50
Thus, the magnitude of acceleration is
A A2T
A2N 1.202
0.06482 1.20 ms2
Ans.
v
12–154.
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
A
x
SOLUTION
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 (v0)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 + (v0)y t +
y = 0 + 5.143t +
1
(ac)y t2
2
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
40
Ans.
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.
y
x
12–155.
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
SOLUTION
B
4 1
at = v4
3
dv = at dt
dv =
4 1
v 4 dt
3
v
L16
3
t
dv
0.75
=
1
v4
L0
dt
v
v4 16 = t
3
v4 - 8 = t
4
v = (t + 8)3
ds = v dt
s
L0
t
ds =
L0
4
(t + 8)3 dt
t
s =
7
3
(t + 8)3 2
7
0
s =
7
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
4
v = (10.108 + 8)3 = 47.551 = 47.6 m>s
at =
an =
Ans.
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.
x
12–156.
A particle P travels along an elliptical spiral path such
that
its
position
vector
r
is
defined
by
r = 52 cos10.1t2i + 1.5 sin10.1t2j + 12t2k6 m, where t is in
seconds and the arguments for the sine and cosine are given
in radians. When t = 8 s, determine the coordinate
direction angles a, b, and g, which the binormal axis to the
osculating plane makes with the x, y, and z axes. Hint: Solve
for the velocity vP and acceleration a P of the particle in
terms of their i, j, k components. The binormal is parallel to
vP * a P . Why?
z
P
r
SOLUTION
y
rP = 2 cos (0.1t)i + 1.5 sin (0.1t)j + 2tk
#
vP = r = - 0.2 sin (0.1t)i + 0.15 cos (0.1t)j + 2k
x
$
aP = r = -0.02 cos 10.1t2i - 0.015 sin 10.1t2j
When t = 8 s,
vP = -0.2 sin (0.8 rad)i + 0.15 cos (0.8 rad)j + 2k = -0.143 47i + 0.104 51j + 2k
aP = -0.02 cos (0.8 rad)i - 0.015 sin (0.8 rad)j = -0.013 934i - 0.010 76 j
Since the binormal vector is perpendicular to the plane containing the n–t axis, and
ap and vp are in this plane, then by the definition of the cross product,
i
b = vP * aP = 3 - 0.14 347
- 0.013 934
j
k
0.104 51 2 3 = 0.021 52i - 0.027 868j + 0.003k
-0.010 76 0
b = 210.0215222 + 1 -0.02786822 + 10.00322 = 0.035 338
ub = 0.608 99i - 0.788 62j + 0.085k
a = cos - 1(0.608 99) = 52.5°
Ans.
b = cos - 1(-0.788 62) = 142°
Ans.
g = cos - 1(0.085) = 85.1°
Ans.
Note: The direction of the binormal axis may also be specified by the unit vector
ub¿ = - ub, which is obtained from b¿ = ap * vp.
For this case, a = 128°, b = 37.9°, g = 94.9°
Ans.
12–157.
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.
SOLUTION
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 v = 262 + 42
= 7.21 m>s. Since the velocity is always directed tangent to the path,
vn = 0
and
vt = 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 =
dv
= {2i + 2j} m>s2
dt
Then
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.
12–158.
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
x2
a2
a
SOLUTION
b
2a2 - x2, we have
a
Acceleration: Differentiating twice the expression y =
dy
bx
= dx
a2a2 - x2
d2y
dx2
= -
ab
(a2 - x2)3>2
The radius of curvature of the path is
c1 + a
r =
2
dy 2 3>2
b d
dx
d2y
dx2
B1 + ¢ =
2
2-
2 3>2
bx
≤ R
a2a2 - x2
ab
(a2 - x2)3>2
c1 +
=
2
3>2
b2x2
d
a2(a2 - x2)
ab
(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 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 B =
4 A
a
a
To determine the normal acceleration, apply Eq. 12–20.
(an)max =
a
v2
v2
= 2 = 2 v2
r
b >a
b
Since the motorcycle is traveling at a constant speed, at = 0. Thus,
amax = (an)max =
a 2
v
b2
Ans.
x
y2
b2
1
12–159
The time rate of change of acceleration is referred to as the jerk, which is often used as a means
of measuring passenger discomfort. Calculate this vector, 1 a , in terms of its cylindrical
2
components, using Eq. 12-32.
Solution:
(
)
a = r − rθ 2 ur + (rθ + 2rθ)uθ + zu Z
(
)
(
)
u + j = a = r − rθ 2 − 2rθθ
r − rθ 2 u r ...
r
rθ + 2rθ)uθ + (rθ + 2rθ)u θ + zu Z + zu z
+(rθ + rθ + 2
ur = θuθ
But
u θ = −θur
u z = 0
Substituting and combining terms yields
(
)
(
)
1
1
1
u + 1 rθ + 3rθ + 3
r − 3rθ 2 − 3rθθ
rθ − rθ3 uθ + (
z )u z
j = r
2
2
2
2
Ans.
12–160.
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.
SOLUTION
r = b - a cos u
#
#
r = a sin uu
$
#
#
r = a cos uu2 + a sin uu
#
#
vr = r = a sin uu
#
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 + 2a a sin uu b u
$
#
= (b - a cos u)u + 2au2 sin u
Ans.
Ans.
Ans.
Ans.
12–161.
If a particle’s position is described by the polar coordinates
r = 411 + sin t2 m and u = 12e -t2 rad, where t is in seconds
and the argument for the sine is in radians, determine the
radial and transverse components of the particle’s velocity
and acceleration when t = 2 s.
SOLUTION
When t = 2 s,
r = 4(1+ sin t) = 7.637
#
r = 4 cos t = -1.66459
$
r = -4 sin t = -3.6372
u = 2 e-t
#
u = - 2 e - t = - 0.27067
$
u = 2 e-t = 0.270665
#
vr = r = -1.66 m>s
#
vu = ru = 7.637(- 0.27067) = - 2.07 m>s
#
$
ar = r - r1u22 = -3.6372 - 7.6371 -0.2706722 = - 4.20 m>s2
$
##
au = ru + 2ru = 7.63710.2706652 + 21- 1.6645921- 0.270672 = 2.97 m>s2
Ans.
Ans.
Ans.
Ans.
12–162. A particle moves along a circular path of radius
300 mm. If its angular velocity is u (2U2) rads, where t is
in seconds, determine the magnitude of the particle’s
acceleration when U 2 s.
Time Derivatives:
S S 0
u 2U2 U 2 s 8 rads
u 4U U 2 s 8 rads2
Velocity: The radial and transverse components of the particle’s velocity are
vS S 0
vu Su 0.3(8) 2.4 ms
Thus, the magnitude of the particle’s velocity is
v 2vS 2
vu 2 202
2.42 2.4 ms
Ans.
Acceleration:
B S S Su2 0 0.3 82 19.2 ms2
B u Su 2Su 0.3(8) 0 2.4 ms2
Thus, the magnitude of the particle’s acceleration is
B 2BS 2
Bu 2 2( 19.2)2
2.42 19.3 ms2
Ans.
12–163.
At the instant shown, the watersprinkler is rotating with an angular speed T' and an angular
acceleration T''. If the nozzle lies in the vertical plane and water is flowing through it at a
constant rate r', determine the magnitudes of the velocity and acceleration of a water particle as it
exits the open end, r.
Given:
T'
2
T''
3
rad
s
rad
2
s
m
s
r'
3
r
0.2 m
Solution:
v
a
2
r' rT'
r T'
2
2
2
rT'' 2r' T'
2
m
s
v
3.03
a
12.63
Ans.
m
2
s
Ans.
12–164.
A
velocity of
$
# radar gun at O rotates with the angular
2
u = 0.1 rad>s and angular acceleration of u = 0.025 rad>s ,
at the instant u = 45°, as it follows the motion of the car
traveling along the circular road having a radius of
r = 200 m. Determine the magnitudes of velocity and
acceleration of the car at this instant.
r ⫽ 200 m
u
O
SOLUTION
Time Derivatives: Since r is constant,
#
$
r = r = 0
Velocity:
#
vr = r = 0
#
vu = ru = 200(0.1) = 20 m>s
Thus, the magnitude of the car’s velocity is
v = 2vr2 + vu2 = 202 + 202 = 20 m>s
Ans.
Acceleration:
#
#
ar = r - ru2 = 0 - 200(0.12) = - 2 m>s2
$
# #
au = r u + 2ru = 200(0.025) + 0 = 5 m>s2
Thus, the magnitude of the car’s acceleration is
a = 2ar2 + au2 = 2(- 2)2 + 52 = 5.39 m>s2
Ans.
12– 165.
A particle is moving along a circular path having a radius r. Its position as a function of time is
given by T= bt2. Determine the magnitude of the particle’s acceleration when T= T1. The particle
starts from rest when T= 0°
Given:
r
b
400 mm
2
rad
T1
2
30 deg
s
Solution:
T
a
T1
t
bt
2
r T'
t
b
T'
2
2
rT''
0.51 s
T''
2b t
2
a
2.32
m
2
s
2b
Ans.
12–166.
If a particle’s position is described by the polar coordinates
r = 12 sin 2u2 m and u = 14t2 rad, where t is in seconds,
determine the radial and transverse components of its
velocity and acceleration when t = 1 s.
SOLUTION
When t = 1 s,
u = 4t = 4
#
u = 4
$
u = 0
r = 2 sin 2u = 1.9787
#
#
r = 4 cos 2u u = -2.3280
$
#
$
r = -8 sin 2u(u)2 + 8 cos 2u u = -126.638
#
vr = r = -2.33 m>s
#
vu = ru = 1.9787(4) = 7.91 m>s
#
$
ar = r - r(u)2 = -126.638 - (1.9787)(4)2 = - 158 m>s2
$
##
au = ru + 2 ru = 1.9787(0) + 2( - 2.3280)(4) = - 18.6 m>s2
Ans.
Ans.
Ans.
Ans.
12–167. If arm OA rotates counterclockwise with a
constant angular velocity of u 2 rads, 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:
S2 4 sin 2u
2SS 8 cos 2uu
4 cos 2uu
S ! ms
S
2 SS
S2 8 2 sin 2uu2
u 2 rads
cos 2u u 4 cos 2uu 2 sin 2uu2 S2
S " ms2
S
u 0
At u 30°,
S u 30° 24 sin 60° 1.861 m
(4 cos 60°)(2)
2.149 ms
S u 30° 1.861
4 0 2 sin 60° 22 (2.149)2
S u 30° 17.37 ms2
1.861
Velocity:
vu Su 1.861(2) 3.722 ms
vS S 2.149 ms
Thus, the magnitude of the peg’s velocity is
v 2vS 2
vu 2 22.1492
3.7222 4.30 ms
Ans.
Acceleration:
B S S Su2 17.37 1.861 22 24.82 ms2
B u S u 2S u 0 2(2.149)(2) 8.597 ms2
Thus, the magnitude of the peg’s acceleration is
B 2BS 2
Bu 2 2( 24.82)2
8.5972 26.3 ms2
Ans.
u
P
The peg moves in the curved slot defined by the
12–168.
lemniscate, and through
the slot in the arm. At u 30°, the
angular
velocity is u 2 rads, and the angular acceleration
is u 1.5 rads2. Determine the magnitudes of the velocity
and acceleration of peg P at this instant.
r2 (4 sin 2 u)m2
r
O
Time Derivatives:
2SS 8 cos 2uu
4 cos 2uu
S A
I ms
S
23 SS
S2 4 8 3 2 sin 2uu
u 2 rads
cos 2uu2 4
4 cos 2uu 2 sin 2uu2 S2
" ms2
S S
u 1.5 rads2
At u 30°,
S u 30° 24 sin 60° 1.861 m
(4 cos 60°)(2)
S u 30° 2.149 ms
1.861
4 cos 60°(1.5) 2 sin 60° 22 (2.149)2
S u 30° 15.76 ms2
1.861
Velocity:
vu Su 1.861(2) 3.722 ms
vS S 2.149 ms
Thus, the magnitude of the peg’s velocity is
v 2BS 2
Bu 2 22.1492
3.7222 4.30 ms
Ans.
Acceleration:
B S S S u2 15.76 1.861 22 23.20 ms2
Bu Su 2S u 1.861(1.5) 2(2.149)(2) 11.39 ms2
Thus, the magnitude of the peg’s acceleration is
B 2BS 2
Bu 2 2( 23.20)2
11.392 25.8 ms2
Ans.
u
P
12–169.
The time rate of change of acceleration is referred to as the
jerk, which is often used as a means of measuring passenger
#
discomfort. Calculate this vector, a, in terms of its
cylindrical components, using Eq. 12–32.
SOLUTION
$
$ #
##
$
a = a r -ru 2 bur + a ru + 2ru buu + z uz
#$
# #
$#
$
###
###
$
##
$
#$
$#
#$
## #
$#
a = ar - ru2 - 2ruu b ur + a r - ru2 bu r + a r u + ru + 2ru + 2r u b uu + a ru + 2ru buu + z uz + zu z
#
#
#
But, ur = uuu uu = - uur
#
uz = 0
Substituting and combining terms yields
#$
#
$#
#
###
#$
$#
###
#
a = a r - 3ru 2 - 3ruu b ur + a 3r u + ru + 3ru - ru3 b uu + az buz
Ans.
12– 170.
The pin follows the path described by the equation r = a + bcos T. At the instant T = T1. the
angular velocity and angular acceleration are T' and T''. Determine the magnitudes of the pin’s
velocity and acceleration at this instant. Neglect the size of the pin.
Given:
a
0.2 m
b
0.15 m
T1
30 deg
T'
0.7
T''
0.5
rad
s
rad
2
s
T
Solution:
T1
r
a b cos T
v
r' rT'
a
r'' rT'
2
2
r'
b sin T T'
2
2
rT'' 2r' T'
2
r''
2
b cos T T' b sin T T''
v
0.237
a
0.278
m
s
m
2
s
Ans.
Ans.
12–171.
The slotted link is pinned
# at O, and as a result of the
constant angular velocity u = 3 rad>s it drives the peg P for
a short distance along the spiral guide r = 10.4u2 m, where
u is in radians. Determine the radial and transverse
components of the velocity and acceleration of P at the
instant u = p>3 rad.
0.5 m
P
r
·
u
r
3 rad/s
SOLUTION
#
u = 3 rad>s
At u =
p
,
3
u
r = 0.4 u
#
#
r = 0.4 u
$
$
r = 0.4 u
O
r = 0.4189
#
r = 0.4(3) = 1.20
$
r = 0.4(0) = 0
#
v = r = 1.20 m>s
#
vu = r u = 0.4189(3) = 1.26 m>s
#
$
ar = r - ru2 = 0 - 0.4189(3)2 = - 3.77 m>s2
$
##
au = r u + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2
Ans.
Ans.
Ans.
Ans.
0.4 u
12–172.
Solve Prob. # 12–171
if the slotted
link has . an angular
#
#
acceleration u = 8 rad>s 2 when u = 3 rad>s at u = p>3 rad.
SOLUTION
#
u = 3 rad/s
u =
0.5 m
r = 0.4 u
#
#
r = 0.4 u
$
$
r = 0.4 u
r
·
u
r
3 rad/s
u
p
3
O
#
u = 3
$
u = 8
r = 0.4189
#
r = 1.20
$
r = 0.4(8) = 3.20
#
vr = r = 1.20 m>s
#
vu = r u = 0.4189(3) = 1.26 m>s
#
$
ar = r - ru2 = 3.20 - 0.4189(3)2 = - 0.570 m>s2
$
##
au = r u + 2 ru = 0.4189(8) + 2(1.20)(3) = 10.6 m>s2
P
Ans.
Ans.
Ans.
Ans.
0.4 u
12–173.
The slotted link is pinned
# at O, and as a result of the
constant angular velocity u = 3 rad>s it drives the peg P for
a short distance along the spiral guide r = 10.4u2 m, where
u is in radians. Determine the velocity and acceleration of
the particle at the instant it leaves the slot in the link, i.e.,
when r = 0.5 m.
0.5 m
P
r
·
u
SOLUTION
u
r = 0.4 u
#
#
r = 0.4 u
$
$
r = 0.4 u
#
u = 3
$
u = 0
O
At r = 0.5 m,
u =
0.5
= 1.25 rad
0.4
#
r = 1.20
$
r = 0
#
vr = r = 1.20 m>s
#
vu = r u = 0.5(3) = 1.50 m>s
#
$
ar = r - r(u)2 = 0 - 0.5(3)2 = - 4.50 m>s2
$
##
au = ru + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2
r
3 rad/s
Ans.
Ans.
Ans.
Ans.
0.4u
12– 174
A truck is traveling along the horizontal# circular curve of radius r with a constant speed v.
Determine the angular rate of rotation u of the radial line r and the magnitude of the truck’s
acceleration.
Given:
r
60 m
v
20
m
s
Solution:
#
u
v
r
a
r T'
#
u
2
a
0.33
6.67
rad
s
m
2
s
Ans.
Ans.
12 –175.
A truck is traveling along the horizontal circular curve of radius r with speed v which is increasing
at the rate v'. Determine the truck’s radial and transverse components of acceleration.
Given:
r
60 m
v
20
v'
3
m
s
m
2
s
Solution:
2
ar
aT
v
r
ar
v'
aT
m
6.67
2
Ans.
s
3.00
m
2
s
Ans.
12–176.
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
SOLUTION
Time Derivatives:
r = 100 cos 2u
#
#
r = ( -200 (sin 2 u)u ) m>s
At u = 15°,
r u = 15° = 100 cos 30° = 86.60 m
#
#
#
r 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.
12–177.
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.
r
(100 cos 2u) m
u
SOLUTION
Time Derivatives:
r = 100 cos 2u
#
#
r = ( - 200 (sin 2u) u ) m>s
$
#
$
r = -200 C (sin 2u) u + 2 (cos 2u) u 2 D m>s2
At u = 15°,
r u = 15° = 100 cos 30° = 86.60 m
#
#
#
r u = 15° = -200 sin 30°u = - 100u m>s
$
#
#
r u = 15° = -200 C sin 30° u + 2 cos 30° u2 D =
$
Velocity: Referring to Fig. a, vr = - 50 cos f and vu = 50 sin f. Thus,
#
vr = r
#
-50 cos f = -100u
and
#
A -100u - 346.41u2 B m>s2
(1)
#
vu = ru
#
50 sin f = 86.60u
(2)
Solving Eqs. (1) and (2) yields
f = 40.89°
#
u = 0.378 rad>s
Ans.
12–178.
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
700 mm
z
O
SOLUTION
u
OA = 2(400) + (300) + (700) = 860.23 mm
2
2
2
x
OB = 2(400) + (300) = 500 mm
2
vr = (200) a
2
500
b = 116 mm>s
860.23
vu = 0
vz = (200) a
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
700
b = 8.14
860.23
Thus, a = { -5.81ur - 8.14uz} mm>s2
Ans.
400 mm
r
y
300 mm
12–179.
A block moves outward along the slot in the platform with
#
a speed of r = 14t2 m>s, where t is in seconds. The platform
rotates at a constant rate of 6 rad/s. If the block starts from
rest at the center, determine the magnitudes of its velocity
and acceleration when t = 1 s.
·
θ = 6 rad/s
r
θ
SOLUTION
#
r = 4t|t = 1 = 4
#
$
u = 6
u = 0
1
L0
dr =
$
r = 4
1
L0
4t dt
r = 2t2 D 10 = 2 m
#
#
v = 3 A r B 2 + A ru B 2 = 2 (4)2 + [2(6)]2 = 12.6 m>s
#
$
$
##
a =
r - ru2 2 + ru + 2 ru 2 =
[4 - 2(6)2 ]2 + [0 + 2(4)(6)]2
= 83.2 m s
2
Ans.
Ans.
12–180.
The rod OA rotates counterclockwise with a constant angular velocity of T'. Two
pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape
is a limaçon described by the equation r = b(c cos(T)). Determine the speed of the slider
blocks at the instant T T1.
Given:
rad
s
T'
5
b
100 mm
c
2
T1
120 deg
Solution:
T
T1
r
b c cos T
r'
b sin T T'
v
2
r' rT'
2
v
1.323
m
s
Ans.
12–181.
The rod OA rotates counterclockwise with a constant angular velocity of T'. Two
pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape
is a limaçon described by the equation r = b(c cos(T)). Determine the acceleration of the
slider blocks at the instant T T1.
Given:
rad
s
T'
5
b
100 mm
c
2
T1
120 deg
Solution:
T
T1
r
b c cos T
r'
b sin T T'
r''
b cos T T'
a
r'' rT'
2
2
2
2r' T'
2
a
8.66
m
2
s
Ans.
12–182.
A cameraman standing at A is following the movement of a
race car, B, which is traveling around a curved track at
# a
constant speed of 30 m>s. Determine the angular rate u at
which the man must turn in order to keep the camera
directed on the car at the instant u = 30°.
vB
BB
r r
2020
mm
AA
r = 2(20) cos u = 40 cos u
#
#
r = -(40 sin u ) u
#
#
v = r u r + r u uu
#
#
v = 2(r)2 + (r u)2
#
#
(30)2 = ( - 40 sin u)2 (u)2 + (40 cos u)2 (u)2
#
(30)2 = (40)2 [sin2 u + cos2 u](u)2
Ans.
2020
mm
uu
2020
mm
SOLUTION
#
30
u =
= 0.75 rad>s
40
30 m/s
2020
mm
12 –183. The slotted arm AB drives pin C through the spiral
groove described by the
equation R A .. If the angular
velocity is constant at ., determine the radial and transverse
components of velocity and acceleration of the pin.
B
C
Time Derivatives: Since . is constant, then . 0.
R A.
R A.
R A. 0
r
.
Velocity: Applying Eq. 12–25, we have
2R R A.
2. R. A..
Acceleration: Applying Eq. 12–29, we have
AR R R.2 0 A..2 A..2
A. R. 2R . 0 2(A.)(.) 2A.2
Ans.
Ans.
Ans.
Ans.
A
12–184. The slotted arm AB drives pin C through the
spiral groove described by the equation R (1.5 .) m, where
. is in radians. If the arm starts from
rest when . 60° and
is driven at an angular velocity of . (4T) rads, 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, . 4 T and . 4 rads2.
R 1.5 .
R 1.5 . 1.5(4T) 6T
R 1.5 . 1.5 (4) 6 ms2
.
Velocity: Integrate the angular rate,
'
)
3
T
D. '0
4TDT, we have . 1
1
Then, R ( (6T2 )) 8 m. At T 1 s, R 6(12)
2
2
and . 4(1) 4 rads. Applying Eq. 12–25, we have
1
(6T2
3
.
)) rad.
) 4.571 m, R 6(1) 6.00 ms
2R R 6.00 ms
2. R. 4.571 (4) 18.3 ms
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
R.2 6 4.571 42 67.1 ms2
A. R . 2R. 4.571(4) 2(6) (4) 66.3 ms2
AR R
Ans.
Ans.
A
12–185. 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 rads, determine the magnitudes of the
velocity and acceleration of peg P at u 30°.
r
O
Time Derivatives:
S2 4 cos 2u
2SS 8 sin 2uu
4 sin 2uu
S A
I ms
S
2 SS
S2 8 sin 2uu
4 sin 2uu
S u 3 rads
2 cos 2u u2 2 cos 2uu2 S2
S
" ms2
u 0
u 3 rads. Thus, when u 30°,
S u 30° 24 cos 60° 22 m
4 sin 60°(3)
S u 30° 7.348 ms
22
40
S u 30° 2 cos 60°(3)2 ( 7.348)2
22
63.64 ms2
Velocity:
vu Su 22(3) 4.243 ms
vS S 7.348 ms
Thus, the magnitude of the peg’s velocity is
v 2vS 2
vu 2 27.3482
( 4.243)2 8.49 ms
Ans.
Acceleration:
BS S Su2 63.64 22(3)2 76.37 ms2
Bu Su 2Su 0 2( 7.348)(3) 44.09 ms2
Thus, the magnitude of the peg’s acceleration is
B 2BS 2
Bu 2 2( 76.37)2
A
P
( 44.09)2 88.2 ms2
Ans.
u
r2 (4 cos 2 u)m2
B
12–186.
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 (3U3/2) rads, where t is in seconds, determine the
magnitudes of the velocity and acceleration of peg P at
u 30°. When U 0, u 0°.
r
O
Time Derivatives:
S2 4 cos 2u
2SS 8 sin 2uu
4 sin 2uu
S A
I ms
S
2 SS
S2 8 sin 2uu
4 sin 2uu
S 2 cos 2u u2 2 cos 2uu2 r 2
r
" ms2
Eu
u 3U32
EU
u
(0°
U
Eu (0
3U32EU
6
u A U52 I rad
5
At u 30° p
rad,
6
6
p
U52
6
5
U 0.7177 s
u 3U32
1.824 rads
U 0.7177 s
9
u U12
2
3.812 rads2
U 0.7177 s
Thus,
S u 30° 24 cos 60° 22 m
4 sin 60°(1.824)
S u 30° 4.468 ms
22
4 sin 60°(3.812)
S u 30° 2 cos 60°(1.824)2 ( 4.468)2
22
A
P
32.86 ms2
u
r2 (4 cos 2 u)m2
B
12–186. Continued
Velocity:
vu Su 22(1.824) 2.579 ms
vS S 4.468 ms
Thus, the magnitude of the peg’s velocity is
v 2vS 2
vu 2 2( 4.468)2
2.5792 5.16 ms
Ans.
Acceleration:
BS S S u2 32.86 22(1.824)2 37.57 ms2
Bu Su 2Su 22(3.812) 2( 4.468)(1.824) 10.91 ms2
Thus, the magnitude of the peg’s acceleration is
B 2BS 2
Bu 2 2( 37.57)2
( 10.91)2 39.1 ms2
Ans.
12–187.
If the circular plate
# rotates clockwise with a constant
angular velocity of u = 1.5 rad>s, determine the magnitudes
of the velocity and acceleration of the follower rod AB
when u = 2>3p rad.
u
A
B
1/2
r ⫽ (10 ⫹ 50 u
SOLUTION
Time Derivaties:
r = A 10 + 50u1>2 B mm
#
#
r = 25u-1>2u mm>s
#
#
1
$
r = 25 c u-1>2u - u-3>2u2 d mm>s2
2
When u =
2p
rad,
3
r|u =
2p
3
= c 10 + 50a
2p 1>2
b d = 82.36 mm
3
2p -1>2
# 2p
r|u = = 25 a b
11.52 = 25.91 mm>s
3
3
1 2p -3>2
$ 2p
r|u = = 25 c 0 - a
b
11.522 d = - 9.279 mm>s2
3
2 3
Velocity: The radial component gives the rod’s velocity.
#
vr = r = 25.9 mm>s
Acceleration: The radial component gives the rod’s acceleration.
#
$
ar = r - ru2 = - 9.279 - 82.36 A 1.52 B = - 195 mm>s2
Ans.
Ans.
) mm
r
12–188.
When u = 2>3p rad, the angular velocity
and angular
#
u
=
1.5
rad>s and
acceleration
of
the
circular
plate
are
$
u = 3 rad>s2, respectively. Determine the magnitudes of the
velocity and acceleration of the rod AB at this instant.
u
A
B
1/2
r ⫽ (10 ⫹ 50 u
SOLUTION
Time Derivatives:
r = (10 + 50u1>2) mm
#
#
r = 25u - 1>2u mm>s
#2
$
1
$
2
r = 25 cu - 1>2 u - u - 3>2u d mm>s
2
When u =
2p
rad,
3
r|u =
2p
3
= c 10 + 50a
2p 1>2
b d = 82.36 mm
3
2p - 1>2
# 2p
r|u = = 25 a b
(1.5) = 25.91 mm>s
3
3
2p - 1>2
1 2p - 3>2
$ 2p
(3) - a b
(1.52) d = 42.55 mm>s2
r|u = = 25 c a b
3
3
2 3
For the rod,
#
v = r = 25.9 mm>s
Ans.
$
a = r = 42.5 mm>s2
Ans.
) mm
r
12–189.
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.
0.5 m
SOLUTION
Velocity: The inclination angle of the ramp is f = tan-1
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 =
ar 2 + au 2 + az 2 =
( -7.264)2 + 02 + 02 = 7.26 m>s2
Ans.
12–190.
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
SOLUTION
Time Derivatives:
r = 0.5 m
#
$
r = r = 0
#
u = A 1.5t2 B rad>s
$
u = (3t) rad>s2
z = 2 - 0.2t2
$
z = - 0.4 m>s2
#
z = ( -0.4t) m>s
When u = 2p rad,
2p = 0.5t3
t = 2.325 s
Thus,
#
u t = 2.325 s = 1.5(2.325)2 = 8.108 rad>s
$
u t = 2.325 s = 3(2.325) = 6.975 rad>s2
#
z t = 2.325 s = - 0.4(2.325) = - 0.92996 m>s
$
z 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 =
ar 2 + au 2 + az 2 =
( -32.867)2 + 3.4872 + ( -0.4)2 = 33.1 m>s2 Ans.
12–191.
For a short distance the train travels along a track having
the shape of a spiral, r = 11000>u2 m, where u is in radians.
If it maintains a constant speed v = 20 m>s, determine the
radial and transverse components of its velocity when
u = 19p>42 rad.
r
u
r
SOLUTION
r =
1000
u
1000 #
#
r = - 2 u
u
Since
#
#
v2 = (r)2 + (r u)2
(20)2 =
(20)2 =
(1000)2 # 2
(1000)2 # 2
(u) +
(u)
4
u
u2
(1000)2
u4
#
(1 + u2)(u)2
Thus,
#
u =
0.02u2
21 + u2
At u =
9p
4
#
u = 0.140
-1000
#
r =
(0.140) = - 2.80
(9p/4)2
#
vr = r = -2.80 m>s
Ans.
#
1000
(0.140) = 19.8 m>s
vu = r u =
(9p/4)
Ans.
1000
u
12–192.
For a short distance the train travels along a track having
the shape of a spiral, r = 11000>u2 m, where u is in radians.
#
If the angular rate is constant, u = 0.2 rad>s, determine the
radial and transverse components of its velocity and
acceleration when u = 19p>42 rad.
r
u
r
SOLUTION
#
u = 0.2
$
u = 0
r =
1000
u
#
#
r = -1000(u - 2) u
#
$
$
r = 2000(u - 3)(u)2 - 1000(u - 2) u
When u =
9p
4
r = 141.477
#
r = -4.002812
$
r = 0.226513
#
vr = r = -4.00 m>s
#
vu = r u = 141.477(0.2) = 28.3 m>s
#
$
ar = r - r(u)2 = 0.226513 - 141.477(0.2)2 = -5.43 m>s2
$
##
au = r u + 2ru = 0 + 2( -4.002812)(0.2) = - 1.60 m>s2
Ans.
Ans.
Ans.
Ans.
1000
u
r (200 100 cos u) mm
12 –193 If the cam
rotates clockwise with a constant
angular velocity of u 5 rads, 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 S (200 100 cos u) mm.
u
Time Derivatives:
100 cos u) mm
S ( 100 sin uu) mms
S 100 sin uu cos uu2 mms2
S (200
u 5 rads
u 0
When u 30°,
S u 30° 200
100 cos 30° 286.60 mm
S u 30° 100 sin 30°(5) 250 mms
S u 30° 100 0
cos 30° 52 2165.06 mms2
Velocity: The radial component gives the rod’s velocity.
vS S 250 mms
Ans.
Acceleration: The radial component gives the rod’s acceleration.
BS S Su2 2156.06 286.60(52) 9330 mms2
Ans.
A
B
12 –194 . At the instant u 30°, the cam rotates with a
clockwise angular
velocity of u 5 rads and and angular
acceleration of u 6 rads2. 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 S (200 100 cos u) mm.
r (200 100 cos u) mm
u
Time Derivatives:
S (200 100 cos u) mm
S ( 100 sin uu) mms
S 100 sin uu cos uu2 mms2
When u 30°,
S u 30° 200
100 cos 30° 286.60 mm
S u 30° 100 sin 30°(5) 250 mms
S u 30° 100 sin 30°(6)
cos 30° 52 2465.06 mms2
Velocity: The radial component gives the rod’s velocity.
vS S 250 mms
Acceleration: The radial component gives the rod’s acceleration.
BS S Su2 2465.06 286.60 52 9630 mms2
Ans.
Ans.
A
B
12 –195. A particle moves along an Archimedean
spiral
R (8.) m , where . is given in radians. If . 4 rads
(constant), determine the radial and transverse components
of the particle’s velocity and acceleration at the instant
. )2 rad. Sketch the curve and show the components on
the curve.
y
r (8.) m
r
Time Derivatives: Since . is constant, . 0.
R 8. 8 2
)
3 4) m
2
R 8 . 8(4) 32.0 ms
.
x
R 8. 0
Velocity: Applying Eq. 12–25, we have
2 R R 32.0 ms
2. R. 4) (4) 50.3 ms
Acceleration: Applying Eq. 12–29, we have
AR R R.2 0 4) 42 201 ms2
A. R. 2R . 0 2(32.0)(4) 256 ms2
2r 32.0 ms
ar 201 ms2
Ans.
Ans.
Ans.
Ans.
2. 50.3 ms
a. 256 ms2
particle has an angular
12 –196. Solve
Prob. 12–195 if the
acceleration . 5 rads2 when . 4 rads at . )2 rad.
y
Time Derivatives: Here,
)
R 8. 82 3 4) m
2
R 8 . 8(5) 40 ms2
r (8.) m
R 8 . 8(4) 32.0 ms
r
.
x
Velocity: Applying Eq. 12–25, we have
2R R 32.0 ms
2. R . 4)(4) 50.3 ms
Ans.
Ans.
2r 32.0 ms
a. 319 ms2
Acceleration: Applying Eq. 12–29, we have
AR R R .2 40 4) 42 161 ms2
A. R . 2R . 4)(5) 2(32.0)(4) 319 ms2
ar 161 ms2
Ans.
Ans.
2. 50.3 ms
12–197.
The partial surface of the cam is that of a logarithmic spiral
r = (40e0.05u) mm, where u is in #radians. If the cam is rotating
at a constant angular rate of u = 4 rad>s, determine the
magnitudes of the velocity and acceleration of the follower
rod at the instant u = 30°.
u
SOLUTION
r = 40e0.05 u
#
#
r = 2e 0.05u u
·
u
# 2
$
$
r = 0.1e 0.0 5 u a u b + 2e 0.05 u u
u =
p
6
#
u = -4
$
u = 0
p
r = 40e 0.05A 6 B = 41.0610
p
#
r = 2e 0.05A 6 B (-4) = -8.2122
p
$
r = 0.1e 0.05A 6 B (-4) 2 + 0 = 1.64244
#
v = r = -8.2122 = 8.21 mm >s
#
$
a = r - r u2 = 1.642 44 - 41.0610(-4) 2 = - 665.33 = -665 mm>s 2
Ans.
Ans.
4 rad/s
r
40e0.05u
12–198.
Solve
if the cam has an angular# acceleration
$ Prob. 12–197,
2
of
u
=
2
rad>s
when
its angular velocity is u = 4 rad>s at
#
u = 30°.
u
SOLUTION
·
r = 40e0.05u
#
#
r = 2e0.05uu
u ⫽ 4 rad/s
#
$
$
r = 0.1e0.05u A u B 2 + 2e0.05uu
u =
p
6
#
u = -4
$
u = -2
p
r = 40e0.05A 6 B = 41.0610
p
#
r = 2e0.05AA 6 B ( -4) = - 8.2122
p
p
$
r = 0.1e0.05A 6 B( -4)2 + 2e0.05A 6 B(- 2) = - 2.4637
#
v = r = 8.2122 = 8.21 mm>s
#
$
a = r - ru 2 = - 2.4637 - 41.0610( -4)2 = - 659 mm>s2
Ans.
Ans.
r ⫽ 40e0.05u
12–199.
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
2 m/s
A
SOLUTION
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 derived.
(sA - sC) + (sB - sC) + sB = l1
(1)
sB + sC = l2
(2)
Eliminating sC from Eqs. (1) and (2) yields
sA + 4sB = l1 = 2l2
Time Derivative: Taking the time derivative of the above equation yields
vA + 4vB = 0
(3)
Since vA = 2 m>s, from Eq. (3)
(+ T)
2 + 4vB = 0
vB = - 0.5 m>s = 0.5 m>s c
Ans.
B
12–200.
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 aD = 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 relative velocity of block A
with respect to block B when this occurs.
D
B
A
d=3m
SOLUTION
For A:
sA + (sA - sC) = l
2vA = vC
2aA = aC = - 3t2
aA = - 1.5t2 = 1.5t2
vA = 0.5t3
:
:
sA = 0.125 t4
:
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)3 = 0.6050
vB = 5(1.0656) = 5.3281 m>s
vA = vB + vA>B
0.6050i = - 5.3281i + vA>B i
vA B = 5.93 m s
:
Ans.
C
12–201.
If the end of the cable at A is pulled down with speed v, determine the speed at which block B
rises.
Given:
v
2
m
s
Solution:
vA
L
v
2sB sA
0
2vB vA
vB
vA
2
vB
1.00
m
s
Ans.
12–202.
Determine the time needed for the load at B to attain a
speed of 8 m>s, starting from rest, if the cable is drawn into
the motor with an acceleration of 0.2 m>s2.
A
vA
SOLUTION
4 sB + sA = l
B
4 nB = - vA
4 aB = - aA
4 aB = -0.2
aB = -0.05 m>s2
(+T)
vB = (vB)0 + aB t
-8 = 0 - (0.05)(t)
t = 160 s
Ans.
vB
Determine the displacement of the log if the
12 –203.
truck at C pulls the cable 1.2 m to the right.
B
S#) L
2S"
(S"
3S"
S# L
3bS"
bS# 0
Since bS# 3bS" bS" 1.2, then
1.2
0.4 m 0.4 m Ans.
C
12–204.
Determine the speed of cylinder A, if the rope is drawn
towards the motor M at a constant rate of 10 m>s.
SOLUTION
Position Coordinates: By referring to Fig. a, the length of the rope written in terms
of the position coordinates sA and sM is
A
3sA + sM = l
h
Time Derivative: Taking the time derivative of the above equation,
A+TB
3vA + vM = 0
Here, vM = 10 m>s. Thus,
3vA + 10 = 0
vA = -3.33 m>s = 3.33 m>s c
Ans.
M
12–205.
If the rope is drawn toward 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.
SOLUTION
Position Coordinates: By referring to Fig. a, the length of the rope written in terms
of the position coordinates sA and sM is
A
3sA + sM = l
h
Time Derivative: Taking the time derivative of the above equation,
A+TB
3vA + vM = 0
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
3
3
t=1 s
Ans.
M
12 –206. If the hydraulic cylinder H draws in rod BC at
1 ms, determine the speed of slider A.
2S(
B
H
S! L
2v( v!
2(1) v!
v! A
2 ms 2 ms Ans.
C
12
–207. If
Ifblock
blockAAis is
moving
downward
a speed
of
12–206.
moving
downward
withwith
a speed
of 4 fts
2while
ms Cwhile
C is moving
up, determine
at 1 ms, determine
the
speed
is moving
up at 2 fts
the speed of
block
B.
of block B.
S!
2S"
S# L
v!
2v"
v# 0
2
2v"
v" 1 0
0.5 ms 0.5 ms C
B
C
A
Ans.
12–207.
moving
downward
at 6 at
fts2 while
block
If block
blockAAis is
moving
downward
ms while
12
–208. If
C is moving
down atdown
18 ftsat
, determine
the speedthe
of block
block
C is moving
6 ms, determine
speedB.of
block B.
S! 2S"
S# L
v! 2v"
v# 0
2
2v"
v" 6 0
4 ms 4 ms C
B
C
A
Ans.
12–209.
If the point A on the cable is moving upwards at vA, determine the speed of block B.
Given:
14
vA
m
s
Solution:
sD sA sD sE
L1
0
2vD vA vE
sD sE sC sE
L2
0
vD vC 2vE
sC sD sC sE
L3
0
2vC vD vE
m
s
Guesses
vC
Given
0
2vD vA vE
0
vD vC 2vE
0
2vC vD vE
§ vC ·
¨ ¸
¨ vD ¸
¨v ¸
© E¹
vB
1
vD
vB
2
m
s
§ vC ·
¨ ¸
¨ vD ¸
¨v ¸
© E¹
Find vC vD vE
vC
1
m
s
vE
1
m
s
§ 2 ·
¨ 10 ¸ m
¨
¸ s
© 6 ¹
Positive means down,
Negative means up
Ans.
12–210.
The motor draws in the cable at C with a constant velocity vC. The motor draws in the cable at
D with a constant acceleration of aD. If vD = 0 when t = 0, determine (a) the time needed for
block A to rise a distance h, and (b) the relative velocity of block A with respect to block B
when this occurs.
Given:
vC
4
aD
8
m
s
m
2
s
h
3m
Solution:
L1
sD 2sA
0
vD 2vA
0
aD 2aA
L2
sB sB sC
0
2vB vC
0
2aB aC
aA
aD
2
vA
aA t
sA
h
vA
aA t
§ t2 ·
¸
©2¹
aA¨
t
vB
2h
aA
1
vC
2
t
vAB
1.225 s
vA vB
Ans.
vAB
2.90
m
s
Ans.
12–211.
Determine the speed of block A if the end of the rope is
pulled down with a speed of 4 m>s.
4 m/s
B
SOLUTION
Position Coordinates: By referring to Fig. a, the length of the cord written in terms
of the position coordinates sA and sB is
sB + sA + 2(sA - a) = l
sB + 3sA = l + 2a
Time Derivative: Taking the time derivative of the above equation,
(+ T )
vB + 3vA = 0
Here, vB = 4 m>s. Thus,
4 + 3vA = 0
vA = - 133 m>s = 1.33 m>s c
Ans.
A
12–212.
The cylinder C is being lifted using the cable and pulley
system shown. If point A on the cable is being drawn toward
the drum with a speed of 2 m>s, determine the speed of the
cylinder.
A
vA
SOLUTION
l = sC + (sC - h) + (sC - h - sA)
C
l = 3sC - 2h - sA
s
0 = 3vC - vA
vC =
vA
-2
=
= - 0.667 m>s = 0.667 m>s c
3
3
Ans.
12–213.
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
A
SOLUTION
Position-Coordinate Equation: Using the Pythagorean theorem to determine lAC,
we have lAC = 2x2A + 82. Thus,
l = lAC + yB
xA
16 = 2x2A + 82 + yB
yB = 16 - 2x2A + 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
yB = yA
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.
12–214.
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 a A = 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
SOLUTION
Position-Coordinate Equation: Using the Pythagorean theorem to determine lAC,
we have lAC = 2x2A + 82. Thus,
l = lAC + yB
xA
16 = 2x2A + 82 + yB
yB = 16 - 2x2A + 64
(1)
dxA
Time Derivative: Taking the time derivative of Eq. (1) Where vA =
and
dt
dyB
vB =
, we have
dt
dyB
xA
dxA
= 2
dt
2xA + 64 dt
xA
vA
vB = 2
2xA + 64
vB =
(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.
v2A = (v0)2A + 2(ac)A C sA - (s0)A D
v2A = 0 + 2(0.2)(8.944 - 0)
vA = 1.891 m>s
Substitute the above results into Eq. (2) yields
vB = -
8.944
28.9442 + 64
(1.891) = -1.41 m>s = 1.41 m>s c
Ans.
Note: The negative sign indicates that velocity vB is in the opposite direction to that
of positive yB.
12 –215. A man can row a boat at 5 ms 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 ms,
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:
vC vS
vCS
yC sin 45°i yC cos 45°j 2j
B
5 cos ui 5 sin uj
Equating i and j component, we have
yC sin 45° 5 cos u
[1]
yC cos 45° 2 5 sin u
[2]
Solving Eqs. [1] and [2] yields
u 28.57°
yC 6.210 ms 6.21 ms
Ans.
Thus, the time t required by the boat to travel from point A to B is
U T"#
2502 502
11.4 s
yC
6.210
Ans.
12 –216. The car is traveling at a constant speed of 100 kmh.
If the rain is falling at 6 ms in the direction shown, determine
the velocity of the rain as seen by the driver.
30
Solution I
1h
km 1000 m
43
43
4
h
1 km
3600 s
27.78 ms.The velocity of the car and the rain expressed in Cartesian vector form
are vD [ 27.78i] ms and vS [6 sin 30°i 6 cos 30°j] [3i 5.196j] ms.
Applying the relative velocity equation, we have
Vector Analysis: The speed of the car is vD 3 100
vS vD
vSD
3i 5.196j 27.78i
vSD
vSD [30.78i 5.196j] ms
Thus, the magnitude of vr/c is given by
vSD 230.782
( 5.196)2 31.2ms
Ans.
and the angle YS/D makes with the x axis is
u tan 1 3
5.196
4 9.58°
30.78
Ans.
Solution II
Scalar Analysis: Referring to the velocity diagram shown in Fig. a and applying the
law of cosines,
vSD 227.782
62 2(27.78)(6) cos 120°
19.91 ms 19.9 ms
Ans.
Using the result of WS/D and applying the law of sines,
sin u
sin 120°
6
31.21
u 9.58°
Ans.
vr
vC 100 km/h
12–217.
The crate C is being lifted by moving the roller at A
downward with a constant speed of vA = 2 m>s along the
guide. Determine the velocity and acceleration of the crate
at the instant s = 1 m. When the roller is at B, the crate
rests on the ground. Neglect the size of the pulley in the
calculation. Hint: Relate the coordinates xC and xA using
the problem geometry, then take the first and second time
derivatives.
4m
B
xA
xC
4m
SOLUTION
s
xC + 2x2A + (4)2 = l
#
#
1
xC + (x2A + 16) - 1/2(2xA)(xA) = 0
2
1
# 2
$
# 2
$
xC - (x2A + 16) - 3/2 (2x2A)(xA ) + (x2A + 16) - 1/2 (xA) + (x2A + 16) - 1/2 (xA)(xA) = 0
2
l = 8 m , and when s = 1 m ,
xC = 3 m
xA = 3 m
#
vA = xA = 2 m>s
$
aA = x A = 0
Thus,
vC + [(3)2 + 16] - 1/2 (3)(2) = 0
vC = - 1.2 m>s = 1.2 m>s c
Ans.
aC - [(3)2 + 16] - 3/2 (3)2(2)2 + [(3)2 + 16] - 1/2 (2)2 + 0 = 0
aC = - 0.512 m>s2 = 0.512 m>s2 c
A
C
Ans.
12–218.
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
5 m/s
u
SOLUTION
A
Solution I
25 m
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,
vb = 5.563 m>s = 5.56 m>s
Ans.
Using the result of vb and applying the law of sines,
sin (180° - u) sin 63.43°
=
5.563
5
u = 84.4°
Ans.
50 m
2 m/s
12 –219. A man walks at 5 kmh in the direction of a
20-kmh wind. If raindrops fall vertically at 7 kmh 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
vS vw
vSw 20i
vm 5 km/h
( 7j) {20i 7j} kmh
Thus, the relative velocity of the rain with respect to the man is
vS vN
20i 7j 5i
vSN
vSN
vSN {15i 7j} kmh
The magnitude of the relative velocity vr/m is given by
ySN 2152
( 7)2 16.6 kmh
Ans.
7
25.0° c
15
Ans.
And its direction is given by
u tan 1
12–220.
If block B is moving down with a velocity vB and has an
acceleration a B , determine the velocity and acceleration of
block A in terms of the parameters shown.
sA
A
h
vB, aB
B
SOLUTION
l = sB + 3s2B + h2
1
#
#
0 = sB + (s2A + h2) - 1/2 2sA sA
2
#
- sB(s2A + h2)1/2
#
vA = sA =
sA
vA = - vB (1 + a
h 2 1/2
b )
sA
Ans.
h 2
h 2
1
#
#
#
aA = vA = - vB(1 + a b )1/2 - vB a b (1 + a b ) - 1/2(h2)( - 2)(sA) - 3sA
sA
2
sA
aA = - aB(1 + a
h 2 1/2 vAvBh2
h 2 - 1/2
b ) +
(1
+
a
b )
sB
sA
s3A
Ans.
12–221.
At a given instant, two particles A and B are moving with a speed of v0 along the paths
shown. If B is decelerating at v'B and the speed of A is increasing at v'A, determine the
acceleration of A with respect to B at this instant.
Given:
v0
a
8
m
s
1m
v'A
5
m
2
s
6
v'B
m
2
s
Solution:
3
2
y ( x)
§x·
a¨ ¸
© a¹
U
1 y' ( a)
y'' ( a)
d
y ( x)
dx
y' ( x)
2
d
y' ( x)
dx
y'' ( x)
3
T
U
atan ( y' ( a) )
2
aA
aAB
§ cos T · v0 § sin T ·
v'A ¨
¸
¨
¸
© sin T ¹ U © cos T ¹
aA aB
aAB
§ 0.2 · m
¨
¸
© 4.46 ¹ s2
aB
aAB
7.812 m
v'B § 1
·
¨ ¸
2 © 1 ¹
4.47
m
2
s
Ans.
12–222.
Two planes, A and B, are flying at the same altitude. If their
velocities are vA = 600 km>h and vB = 500 km>h such that
the angle between their straight-line courses is u = 75°,
determine the velocity of plane B with respect to plane A.
A
vA
B
u
SOLUTION
vB
vB = vA + vB/A
75°
[500 ; ] = [600 c u ] + vB/A
+ )
(;
500 = -600 cos 75° + (vB/A)x
(vB/A)x = 655.29 ;
(+ c)
0 = -600 sin 75° + (vB/A)y
(vB/A)y = 579.56 c
1vB>A2 = 31655.2922 +1579.5622
vB/A = 875 km/h
u = tan-1 a
579.56
b = 41.5° b
655.29
Ans.
Ans.
12
–223. If the truck travels at a constant speed of
12–214.
vVT4 61.8
determine
the speed
the for
crate
any.
, determine
the speed
of theofcrate
anyfor
angle
ftsms,
angle
of the
hasof a100
length
30 m over
and
of the .rope.
Therope.
rope The
has arope
length
ft andofpasses
passes
a pulley of
negligible
sizeRelate
at A. Hint:
Relate the
a pulleyover
of negligible
size
at A. Hint:
the coordinates
coordinates
xT and
of theand
rope
andthe
taketime
the
the xlength
oflength
the rope
take
X4 and X# to
C to the
time
derivative.
substitute
trigonometric relation
derivative.
ThenThen
substitute
thethetrigonometric
between xXC# and ..
..
(6)2
X2#
1
(6)2
2
(X#)2 xC
6m
C
.
X4 L 30
1
2
2 2X#X# 3
X4 0
Since X4 v4 1.8 ms, vc xc, and
X# 6 ctn .
Then,
(6 ctn .)v#
(36
Since 1
2
1
36 ctn 2 .) 2
1.8
ctn2 . csc2 .,
ctn .
3v# cos .v# csc .
v# 6m
1.8
1.8 sec . (1.8 sec .) ms xT
A
Ans.
T
vT
12–224.
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
A
SOLUTION
200 m
C
30
Velocity: Referring to Fig. a, the velocity of cars A and B expressed in Cartesian
vector form are
vA = [25 cos 30° i - 25 sin 30° j] m>s = [21.65i - 12.5j] m>s
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
vB>A = 2( -7.162)2 + 8.6182 = 11.2 m>s
Ans.
The direction angle uv of vB/A measured down from the negative x axis, Fig. b is
uv = tan - 1 a
8.618
b = 50.3° d
7.162
Ans.
B
15
12–225.
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
SOLUTION
50 km/h
vB = vC + vB>C
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
45.293
b = 67.2° d
19.04
A
Ans.
Ans.
12–226.
A car is traveling north along a straight road at 50 km>h.
An instrument in the car indicates that the wind is directed
toward the east. If the car’s speed is 80 km>h, the
instrument indicates that the wind is directed toward the
north-east. Determine the speed and direction of the wind.
SOLUTION
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
vw = vc + 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.
12 –227. The motor at C pulls in the cable with an
acceleration B$ (3U2) ms2, where t is in seconds. The
motor at D draws in its cable at B% 5 ms2. If both motors
start at the same instant from rest when E 3 m, determine
(a) the time needed for E 0, and (b) the velocities of
blocks A and B when this occurs.
D
B
A
d3m
For A:
T"
(T" T$) M
2v" v$
2B" B$ 3U2
B" 1.5U2 1.5U2 v" 0.5U3 T" 0.125U4 For B:
B# 5 ms2 v# 5U T# 2.5U2 Require T"
0.125U4
Set V U2
T# E
2.5U2 3
0.125V2
2.5V 3
The positive root is V 1.1355. Thus,
U 1.0656 1.07 s
Ans.
v" .0.5(1.0656)2 0.605 ms
Ans.
v# 5(1.0656) 5.33 ms
Ans.
C
12–228.
At the instant shown, the bicyclist at A is traveling at 7 m/s
around the curve on the race track while increasing his speed
at 0.5 m>s2. The bicyclist at B is traveling at 8.5 m/s along the
straight-a-way and increasing his speed at 0.7 m>s2. Determine
the relative velocity and relative acceleration of A with respect
to B at this instant.
vB = 8.5 m/s
B
50 m
vA = vB + vA>B
[7 |R ] = [8.5 : ] + [(vA>B)x : ] + [(vA>B)y T]
40°
7 sin 40° = 8.5 + (vA>B)x
(+ T )
7 cos 40° = (vA>B)y
Thus,
(vA>B)x = 4.00 m>s ;
(vA>B)y = 5.36 m>s T
(vA>B) = 2(4.00)2 + (5.36)2
vA>B = 6.69 m>s
u = tan - 1 a
(aA)n =
Ans.
5.36
b = 53.3° d
4.00
Ans.
72
= 0.980 m>s2
50
aA = aB + aA>B
[0.980] ud
40° + [0.5] |R
40°
(+ : )
= [0.7 : ] + [(aA>B)x : ] + [(aA>B)y T]
- 0.980 cos 40° + 0.5 sin 40° = 0.7 + (aA>B)x
(aA>B)x = 1.129 m>s2 ;
(+ T )
0.980 sin 40° + 0.5 cos 40° = (aA>B)y
(aA>B)y = 1.013 m>s2 T
(aA>B) = 2(1.129)2 + (1.013)2
aA>B = 1.52 m>s2
u = tan - 1
1.013
1.129
Ans.
= 41.9° d
50 m
40°
SOLUTION
+ )
(:
vA = 7 m/s
Ans.
A
12–229.
At the instant shown car A is traveling with a velocity vA and has an acceleration aA along the
highway. At the same instant B is traveling on the trumpet interchange curve with a speed vB
which is decreasing at v'B. Determine the relative velocity and relative acceleration of B with
respect to A at this instant.
Given:
vA
30
m
s
vB
15
m
s
aA
2
m
2
s
0.8
v'B
m
2
s
U
250 m
T
60 deg
Solution:
vAv
§ vA ·
¨ ¸
©0¹
vBv
§ cos T ·
vB¨
¸
© sin T ¹
aAv
§ aA ·
¨ ¸
©0¹
2
aBv
§ cos T · vB § sin T ·
v'B ¨
¸
¨
¸
© sin T ¹ U © cos T ¹
vBA
vBv vAv
vBA
§ 22.5 · m
¨
¸
© 12.99 ¹ s
vBA
26.0
aBA
aBv aAv
aBA
§ 1.621 · m
¨
¸
© 1.143 ¹ s2
aBA
1.983
m
s
Ans.
m
2
s
Ans.
12–230.
The two cyclists A and B travel at the same constant speed v.
Determine the speed 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
f
B
v
SOLUTION
vA = v sin ui + v cos uj
vB = vi
vA>B = vA - vB
= (v sin ui + v cos uj) - vi
= (v sin u - v)i + v cos uj
vA>B = 2(v sin u - v)2 + (v cos u)2
= 22v2 - 2v2 sin u
= v 22(1 - sin u)
Ans.
u
12 –231. At the instant shown, cars A and B travel at
*12–224.
At the instant shown, cars A and B travel at speeds
speeds of 70 kmh and 50 kmh, respectively. If B is
of 70 mih and 50 mih, respectively. If2 B is increasing its speed
increasing its speed by 1100 kmh , while A maintains a
by 1100 mih2, while A maintains a constant speed, determine
constant speed, determine the velocity and acceleration of
the velocity and acceleration of B with respect to A. Car B
B with respect to A. Car B moves along a curve having a
moves along a curve having a radius of curvature of 0.7 mi.
radius of curvature of 0.7 km.
A
VA 70 kmh
Relative Velocity:
v" v!
v"!
30°
50 cos 30°j 70j
50 sin 30°i
v"! {25.0i
v"!
26.70j} kmh
Thus, the magnitude of the relative velocity vB/A is
2"! 25.02
( 26.70)2 36.6 kmh
Ans.
The direction of the relative velocity is the same as the direction of that for relative
acceleration. Thus
1
. tan
26.70
46.9° 25.0
Ans.
Relative Acceleration: Since car B is traveling along a curve, its normal
22"
502
acceleration is (A")N 3571.43 kmh2. Applying Eq. 12–35 gives
+
0.7
a" a!
(1100 sin 30°
3571.43 cos 30°)i
a"!
(1100 cos 30°
a"! {3642.95i
3571.43 sin 30°)j 0
a"!
833.09j} kmh2
Thus, the magnitude of the relative velocity aB/A is
A"! 3642.952
( 833.09)2 3737 kmh2
Ans.
And its direction is
tan
1
833.09
12.9° 3642.95
VB 50 kmh
Ans.
B
•12–225.
12 –232. At the instant shown, cars A and B travel at
speeds of 70 kmh
mih and 50 kmh,
mih, respectively.
respectively. If B is
decreasing its speed
speed by
at 1400 kmh
mih22,while
while A
A is
is increasing its
speed at 800 mih
kmh22,, determine
determine the
the acceleration
acceleration of B with
respect to A. Car B moves along a curve having a radius of
curvature of 0.7 mi.
km.
A
VA 70 kmh
Relative Acceleration: Since car B is traveling along a curve, its normal acceleration
22"
502
is (A")N 3571.43 kmh2. Applying Eq. 12–35 gives
+
0.7
a" a!
(3571.43 cos 30°
1400 sin 30°)i
a"!
( 1400 cos 30°
a"! {2392.95i
3571.43 sin 30°)j 800j
a"!
3798.15j} kmh2
Thus, the magnitude of the relative acc. aB/A is
A"! 2392.952
( 3798.15)2 4489 kmh2
Ans.
And its direction is
tan
1
3798.15
57.8° 2392.95
Ans.
VB 50 kmh
30°
B
12–233.
A passenger in an automobile observes that raindrops make
an angle of 30° with the horizontal as the auto travels
forward with a speed of 60 km/h. Compute the terminal
(constant)velocity vr of the rain if it is assumed to fall
vertically.
vr
va = 60 km/h
SOLUTION
vr = va + vr>a
-vr j = - 60i + vr>a cos 30°i - vr>a sin 30°j
+ )
(:
0 = - 60 + vr>a cos 30°
(+ c )
- vr = 0 - vr>a sin 30°
vr>a = 69.3 km>h
vr = 34.6 km h
Ans.
12–234.
At the instant shown, the car at A is traveling at vA around the curve while increasing its speed
at v'A. The car at B is traveling at vB along the straightaway and increasing its speed at v'B.
Determine the relative velocity and relative acceleration of A with respect to B at this instant.
Given:
vA
10
v'A
5
m
s
m
2
vB
18.5
v'B
2
s
T
45 deg
m
s
m
2
s
U
100 m
Solution:
§ sin T ·
¸
© cos T ¹
vAv
vA¨
aAv
§ sin T · vA § cos T ·
v'A ¨
¸
¨
¸
© cos T ¹ U © sin T ¹
vBv
§ vB ·
¨ ¸
©0¹
vAB
vAv vBv
vAB
§ 11.43 · m
¨
¸
© 7.07 ¹ s
Ans.
aAB
aAv aBv
aAB
§ 0.828 · m
¨
¸
© 4.243 ¹ s2
Ans.
2
aBv
§ v'B ·
¨ ¸
© 0 ¹
12–235.
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
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 = 19.9 m>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° d
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° d
Ans.
12–236.
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
Velocity: The velocity of cars A and C 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
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 C 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
4.061
b = 75.4° d
1.061
Ans.
3 m/s2
30
15 m/s
SOLUTION
uv = tan - 1 a
45
2
2 m/s2
vA>C = 2( -17.68)2 + 12.322 = 21.5 m>s
A
25 m/s
B
C
30 m/s
12–237.
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.
r
100 m
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
Ans.
and the direction angle uv that vB/C makes with the x axis is
17.01
b = 66.2° a
7.5
Acceleration: The normal component of car B’s acceleration is (aB)n =
Ans.
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
0.1429
b = 8.57° c
0.9486
Ans.
3 m/s2
30
15 m/s
SOLUTION
uv = tan - 1 a
45
1.5 m/s2
2 m/s2
vB>C = 27.52 + 17.012 = 18.6 m>s
A
25 m/s
B
C
30 m/s
12–238.
At a given instant the football player at A throws a football C
with a velocity of 20 m/s in the direction shown. Determine
the constant speed at which the player at B must run so that
he can catch the football at the same elevation at which it was
thrown. Also calculate the relative velocity and relative
acceleration of the football with respect to B at the instant the
catch is made. Player B is 15 m away from A when A starts to
throw the football.
A
15 m
Ball:
+ )s = s + v t
(:
0
0
sC = 0 + 20 cos 60° t
v = v0 + ac t
- 20 sin 60° = 20 sin 60° - 9.81 t
t = 3.53 s
sC = 35.31 m
Player B:
+ ) s = s + n t
(:
B
B
0
Require,
35.31 = 15 + vB (3.53)
vB = 5.75 m>s
Ans.
At the time of the catch
(vC)x = 20 cos 60° = 10 m>s :
(vC)y = 20 sin 60° = 17.32 m>s T
vC = vB + vC>B
10i - 17.32j = 5.751i + (vC>B)x i + (vC>B)y j
+ )
(:
10 = 5.75 + (vC>B)x
(+ c)
- 17.32 = (vC>B)y
(vC>B)x = 4.25 m>s :
(vC>B)y = 17.32 m>s T
vC>B = 2(4.25)2 + (17.32)2 = 17.8 m>s
Ans.
17.32
b = 76.2°
4.25
Ans.
u = tan - 1 a
c
aC = aB + aC>B
- 9.81 j = 0 + aC>B
aC B = 9.81 m s2 T
60°
B
SOLUTION
(+ c )
C
20 m/s
Ans.
12–239.
Both boats A and B leave the shore at O at the same time. If
A travels at vA and B travels at vB, write a general
expression to determine the velocity of A with respect to B.
A
B
SOLUTION
O
Relative Velocity:
vA = vB + vA>B
vA j = vB sin ui + vB cos uj + vA>B
vA>B = - vB sin ui + (vA - vB cos u)j
Thus, the magnitude of the relative velocity vA>B is
vA>B = 2( - vB sin u)2 + (vA - vB cos u)2
= 2vA2 + vB2 - 2vA vB cos u
Ans.
And its direction is
u = tan-1
vA - vB cos u
vB sin u
b
Ans.