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12–1.
Starting from rest, a particle moving in a straight line has an
acceleration of a = (2t - 6) m>s2, where t is in seconds. What
is the particle’s velocity when t = 6 s, and what is its position
when t = 11 s?
Solution
a = 2t - 6
dv = a dt
L0
v
dv =
L0
t
(2t - 6) dt
v = t 2 - 6t
ds = v dt
L0
s
ds =
s =
L0
t
(t2 - 6t) dt
t3
- 3t2
3
When t = 6 s,
Ans.
v = 0
When t = 11 s,
Ans.
s = 80.7 m
@solutionmanual1
1
Ans:
v = 0
s = 80.7 m
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12–2.
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.
Ans:
v = 625 m>s
s = 639.5 m
d = 637.5 m
@solutionmanual1
2
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12–3.
The velocity of a particle traveling in a straight line is given
by v = (6t - 3t2) m>s, where t is in seconds. If s = 0 when
t = 0, determine the particle’s deceleration and position
when t = 3 s. How far has the particle traveled during the
3-s time interval, and what is its average speed?
Solution
v = 6t - 3t 2
a =
dv
= 6 - 6t
dt
At t = 3 s
a = - 12 m>s2
Ans.
ds = v dt
L0
s
ds =
L0
t
(6t - 3t2)dt
s = 3t 2 - t 3
At t = 3 s
Ans.
s = 0
Since v = 0 = 6t - 3t 2, when t = 0 and t = 2 s.
when t = 2 s, s = 3(2)2 - (2)3 = 4 m
Ans.
sT = 4 + 4 = 8 m
( vsp ) avg =
sT
8
= = 2.67 m>s
t
3
Ans.
Ans:
a = - 12 m>s2
s = 0
sT = 8 m
(vsp)avg = 2.67 m>s
@solutionmanual1
3
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*12–4.
A particle is moving along a straight line such that its
position is defined by s = (10t2 + 20) mm, where t is in
seconds. Determine (a) the displacement of the particle
during the time interval from t = 1 s to t = 5 s, (b) the
average velocity of the particle during this time interval,
and (c) the acceleration when t = 1 s.
SOLUTION
s = 10t2 + 20
(a) s|1 s = 10(1)2 + 20 = 30 mm
s|5 s = 10(5)2 + 20 = 270 mm
¢s = 270 - 30 = 240 mm
Ans.
(b) ¢t = 5 - 1 = 4 s
vavg =
(c) a =
240
¢s
=
= 60 mm>s
¢t
4
d2s
= 20 mm s2
dt2
Ans.
(for all t)
Ans.
Ans:
∆s = 240 mm
vavg = 60 mm>s
a = 20 mm>s2
@solutionmanual1
4
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12–5.
A particle moves along a straight line such that its position
is defined by s = (t2 - 6t + 5) m. Determine the average
velocity, the average speed, and the acceleration of the
particle when t = 6 s.
Solution
s = t2 - 6t + 5
v =
ds
= 2t - 6
dt
a =
dv
= 2
dt
v = 0 when t = 3
s t=0 = 5
s t = 3 = -4
s t=6 = 5
vavg =
∆s
0
= = 0
∆t
6
( vsp ) avg =
Ans.
sT
9 + 9
=
= 3 m>s
∆t
6
Ans.
a t = 6 = 2 m>s2
Ans.
Ans:
vavg = 0
(vsp)avg = 3 m>s
@solutionmanual1
5
a t = 6 s = 2 m>s2
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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.
Ans:
s = 14.71 m
@solutionmanual1
6
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12–7.
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.
Ans:
t = 25 s
s = 312.5 m
@solutionmanual1
7
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*12–8.
A particle travels along a straight line with a velocity
v = (12 - 3t 2) m>s, where t is in seconds. When t = 1 s, the
particle is located 10 m to the left of the origin. Determine
the acceleration when t = 4 s, the displacement from
t = 0 to t = 10 s, and the distance the particle travels during
this time period.
SOLUTION
v = 12 - 3t 2
(1)
dv
= - 6t
dt
a =
s
L-10
ds =
L1
t=4
t
v dt =
= -24 m>s2
L1
t
Ans.
( 12 - 3t 2 ) dt
s + 10 = 12t - t 3 - 11
s = 12t - t 3 - 21
s
t=0
s
t = 10
= - 21
= -901
∆s = -901 - ( -21) = -880 m
Ans.
From Eq. (1):
v = 0 when t = 2s
s
t=2
= 12(2) - (2)3 - 21 = - 5
Ans.
sT = (21 - 5) + (901 - 5) = 912 m
Ans:
a = - 24 m>s2
∆s = - 880 m
sT = 912 m
@solutionmanual1
8
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12–9.
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
B
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
Ans.
Ans:
@solutionmanual1
9
S BA = `
2vA vB - v2A
`
2aA
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12–10.
A particle travels along a straight-line path such that in 4 s
it moves from an initial position sA = - 8 m to a position
sB = +3 m. Then in another 5 s it moves from sB to
sC = -6 m. Determine the particle’s average velocity and
average speed during the 9-s time interval.
SOLUTION
Average Velocity: The displacement from A to C is ∆s = sC - SA = -6 - ( - 8)
= 2 m.
∆s
2
vavg =
=
= 0.222 m>s
Ans.
∆t
4 + 5
Average Speed: The distances traveled from A to B and B to C are sA S B = 8 + 3
= 11.0 m and sB S C = 3 + 6 = 9.00 m, respectively. Then, the total distance traveled
is sTot = sA S B + sB S C = 11.0 + 9.00 = 20.0 m.
(vsp)avg =
sTot
20.0
=
= 2.22 m>s
∆t
4 + 5
Ans.
Ans:
vavg = 0.222 m>s
(vsp)avg = 2.22 m>s
@solutionmanual1
10
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12–11.
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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.
Ans:
t = 30 s
s = 792 m
@solutionmanual1
11
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*12–12.
A particle moves along a straight line with an acceleration
of a = 5>(3s1>3 + s 5>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 a numerical method to evaluate
the integral.
SOLUTION
a =
5
1
3
5
A 3s + s2 B
a ds = v dv
2
v
5 ds
1
3
L1 A 3s + s
0.8351 =
5
2
B
=
L0
v dv
1 2
v
2
v = 1.29 m>s
Ans.
Ans:
v = 1.29 m>s
@solutionmanual1
12
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12–13.
The acceleration of a particle as it moves along a straight
line is given by a = (2t - 1) m>s2, where t is in seconds. If
s = 1 m and v = 2 m>s when t = 0, determine the
particle’s velocity and position when t = 6 s. Also,
determine the total distance the particle travels during this
time period.
Solution
a = 2t - 1
dv = a dt
L2
v
t
L0
dv =
(2t - 1)dt
v = t2 - t + 2
dx = v dt
Lt
s
ds =
s =
L0
t
(t2 - t + 2)dt
1 3
1
t - t 2 + 2t + 1
3
2
When t = 6 s
v = 32 m>s
Ans.
s = 67 m
Ans.
Since v ≠ 0 for 0 … t … 6 s, then
Ans.
d = 67 - 1 = 66 m
Ans:
v = 32 m>s
s = 67 m
d = 66 m
@solutionmanual1
13
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12–14.
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
at
2 c
s3 = 27 900 + 30(30) +
1
(- 1)(302)
2
= 28 350 m = 28.4 km
Ans.
Ans:
s = 28.4 km
@solutionmanual1
14
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12–15.
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
Ans.
16(2)(8t + 1)(8)
dv
256
=
=
m>s2
dt
(8t + 1)4
(8t + 1)3
Ans.
Ans:
16
m>s
(8t + 1)2
256
a=
m>s2
(8t + 1)3
v =
@solutionmanual1
15
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*12–16.
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.
Ans:
t = 48.3 s
@solutionmanual1
16
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12–17.
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
dn
= - kn3
dt
a =
t
n
n - 3 dn =
Ln0
L0
- k dt
1 -2
1n - n0- 22 = - kt
2
-
n = a 2kt + a
1
-2
1
b
b
n20
Ans.
ds = n dt
s
L0
s =
s =
t
ds =
L0
dt
a2kt + a
2a 2kt + a
2k
1
2
1
bb
v20
1
t
2
1
b
b
3
n20
0
1
1
1
B ¢ 2kt + ¢ 2 ≤ ≤ - R
n0
k
n0
1
2
Ans.
Ans:
v = a2kt +
s=
@solutionmanual1
17
1 - 1>2
b
v20
1
1 1>2
1
c a2kt + 2 b d
k
v0
v0
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12–18.
A particle is moving along a straight line with an initial
velocity of 6 m>s when it is subjected to a deceleration of
a = (- 1.5v1>2) m>s2, where v is in m>s. Determine how far it
travels before it stops. How much time does this take?
SOLUTION
Distance Traveled: The distance traveled by the particle can be determined by
applying Eq. 12–3.
ds =
vdv
a
s
L0
v
ds =
v
1
L6 m>s - 1.5v2
v
s =
dv
1
L6 m>s
- 0.6667 v2 dv
3
= a -0.4444v2 + 6.532 b m
When v = 0,
3
s = - 0.4444a 0 2 b + 6.532 = 6.53 m
Ans.
Time: The time required for the particle to stop can be determined by applying
Eq. 12–2.
dt =
dv
a
t
L0
v
dt = 1
t = - 1.333av2 b
When v = 0,
v
6 m>s
dv
1
L6 m>s 1.5v 2
1
= a3.266 - 1.333v 2 b s
1
t = 3.266 - 1.333 a 0 2 b = 3.27 s
Ans.
Ans:
s = 6.53 m
t = 3.27 s
@solutionmanual1
18
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12–19. The acceleration of a rocket traveling upward is
given by a = (6 + 0.02s) m>s2, where s is in meters. Determine
the rocket’s velocity when s = 2 km and the time needed to
reach this attitude. Initially, v = 0 and s = 0 when t = 0.
SOLUTION
b
6 m>s2
c
ap
b c sp
vp
´
µ
¶
vp
vp dvp
0
´
µ
¶
sp1
2000 m
dvp
s
dsp
sp
b c sp dsp
0
2
vp
b sp 2
dsp
vp
t
0.02 s- 2
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
vp1
1
2
dsp
t1
322.49 m>s
Ans.
19.27 s
Ans.
2b sp c sp
0
Ans:
vp1 = 322.49
t1 = 19.27 s
@solutionmanual1
19
m
s
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*12–20.
The acceleration of a rocket traveling upward is given by
a = 16 + 0.02s2 m>s2, where s is in meters. Determine the
time needed for the rocket to reach an altitude of
s = 100 m. Initially, v = 0 and s = 0 when t = 0.
SOLUTION
a ds = n dv
s
s
L0
16 + 0.02 s2 ds =
6 s + 0.01 s2 =
n
L0
n dn
1 2
n
2
n = 212 s + 0.02 s2
ds = n dt
100
L0
ds
212 s + 0.02 s2
1
20.02
t
=
L0
dt
1n B 212s + 0.02s2 + s20.02 +
t = 5.62 s
12
2 20.02
R
100
= t
0
Ans.
Ans:
t = 5.62 s
@solutionmanual1
20
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12–21.
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.
v
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.
Ans:
v = 3.93 m>s
s = 9.98 m
@solutionmanual1
21
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12–22.
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.
Ans:
(a) s = - 30.5 m
(b) sTot = 56.0 m
(c) v = 10 m>s
@solutionmanual1
22
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12–23.
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)
L0 2(1 - 0.01v)
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.
Ans:
(a) v = 45.5 m>s
(b) v max = 100 m>s
@solutionmanual1
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*12–24.
A sandbag is dropped from a balloon which is ascending
vertically at a constant speed of 6 m>s. If the bag is released
with the same upward velocity of 6 m>s when t = 0 and hits
the ground when t = 8 s, determine the speed of the bag as
it hits the ground and the altitude of the balloon at this
instant.
SOLUTION
(+ T )
s = s0 + v0 t +
1
a t2
2 c
h = 0 + ( -6)(8) +
1
(9.81)(8)2
2
= 265.92 m
During t = 8 s, the balloon rises
h¿ = vt = 6(8) = 48 m
Altitude = h + h¿ = 265.92 + 48 = 314 m
(+ T)
Ans.
v = v0 + ac t
v = - 6 + 9.81(8) = 72.5 m s
Ans.
Ans:
h = 314 m
v = 72.5 m>s
@solutionmanual1
24
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12–25.
A particle is moving along a straight line such that its
acceleration is defined as a = (-2v) m>s2, where v is in
meters per second. If v = 20 m>s when s = 0 and t = 0,
determine the particle’s position, velocity, and acceleration
as functions of time.
Solution
a = - 2v
dv
= - 2v
dt
v
t
dv
-2 dt
=
L20 v
L0
ln
v
= -2t
20
v = ( 20e -2t ) m>s
a =
L0
dv
=
dt
Ans.
( - 40e -2t ) m>s2
s
ds = v dt =
L0
Ans.
t
(20e-2t)dt
s = - 10e -2t t0 = - 10 ( e -2t - 1 )
s = 10 ( 1 - e -2t ) m
Ans.
Ans:
v = ( 20e -2t ) m>s
a = ( - 40e -2t ) m>s2
s = 10 ( 1 - e -2t ) m
@solutionmanual1
25
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12–26.
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 =
v
1 1>2
s ds
L1 4
s
v2
2 = 1 s3>2 `
2 0
6
1
v =
1
23
1s3>2 - 121>2 m>s
When s = 2 m, v = 0.781 m>s.
Ans.
Ans:
v = 0.781 m>s
@solutionmanual1
26
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12–27.
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
2¿
- v
=
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.
Ans:
@solutionmanual1
27
t = 0.549 a
vf
g
b
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*12–28.
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.
Ans:
s = 54.0 m
@solutionmanual1
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12–29.
A ball A is thrown vertically upward from the top of a
30-m-high building with an initial velocity of 5 m>s. At the
same instant another ball B is thrown upward from
the ground with an initial velocity of 20 m>s. Determine the
height from the ground and the time at which they pass.
Solution
Origin at roof:
Ball A:
1
2
( + c ) s = s0 + v0t + act 2
- s = 0 + 5t -
1
(9.81)t 2
2
Ball B:
1
2
( + c ) s = s0 + v0t + act 2
- s = -30 + 20t -
1
(9.81)t 2
2
Solving,
Ans.
t = 2 s
s = 9.62 m
Distance from ground,
Ans.
d = (30 - 9.62) = 20.4 m
Also, origin at ground,
s = s0 + v0t +
1 2
at
2 c
sA = 30 + 5t +
1
( -9.81)t 2
2
sB = 0 + 20t +
1
( - 9.81)t 2
2
Require
sA = sB
30 + 5t +
1
1
( -9.81)t 2 = 20t + ( -9.81)t 2
2
2
t = 2 s
Ans.
sB = 20.4 m
Ans.
@solutionmanual1
29
Ans:
h = 20.4 m
t = 2s
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12–30.
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.
@solutionmanual1
30
Ans:
v1 = 3.68 m>s
t2 = 1.98 s
v2 = 15.8 m>s T