SOLUTION
a=2t–6
dv=a dt
Lv
0
dv=
Lt
0
(2t –6) dt
v=t2–6t
ds =v dt
Ls
0
ds=
Lt
0
(t2–6t) dt
s=
t
3
3
–3t2
When
t=6 s
,
v=0
Ans.
When
t=11 s
,
s=80.7 m
Ans.
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?
12–2.
SOLUTION
If a particle has an initial velocity of v0=12 ft>s to the
right, at s0=0, determine its position when t=10 s, if
a=2 ft>s2 to the left.
12–3.
A particle travels along a straight line with a velocity
v=(12–3t2) m>s, where t is in seconds. When t=1 s, the
particle is located 10 m to the left of the origin. Determine
the acceleration when t=4
s, the displacement from
t=0 to t=10 s, and the distance the particle travels during
this time period.
4
*12–4.
SOLUTION
and and apply Eq. 12–6.
Using the result , the velocity function can be obtained by applying
Eq. 12–6.
v2=32+2(4.583) (s–4)
(:
+)v2=v2
0+2ac(s–s0)
ac=4.583 ft>s2
ac=4.583 ft>s2
82=32+2ac(10 –4)
(:
+)v2=v2
0+2ac(s–s0)
v=8ft>ss=10 ft
Apart
i
c
l
e trave
l
s a
l
ong a stra
i
g
h
t
li
ne w
i
t
h
a constant
acceleration. When ,and when ,
.Determine the velocity as a function of position.v=8ft>s
s=10 ftv=3ft>ss=4ft
5
12–5.
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-stime interval, and what is its average speed?
SOLUTION
v=6t–3t2
a=
dv
dt
=6–6t
At
t=3 s
a
=–
12 m
>
s
2
Ans.
ds =v dt
L
s
0
ds =
Lt
0
(6t –3t2)dt
s=3t2–t3
At
t=3 s
s=0
Ans.
Since
v=0=6
t
–3
t
2,
when
t=0
and
t=2 s.
when
t=2 s
, s
=
3(2)
2–
(2)
3=
4 m
sT=4+4=8 m
avg
t
3
Ans:
sT=8 m
vavg
=2.67 m>s
12–6.
SOLUTION
Ans.
Total Distance Traveled:The velocity of the particle can be determined by applying
Eq. 12–1.
The times when the particle stops are
The position of the particle at ,1 s and 5 s are
From the particle’s path, the total distance is
Ans.stot =10.5 +48.0 +10.5 =69.0 ft
st=5 s =1.5(53)–13.5(52)+22.5(5) =-37.5 ft
st=1 s =1.5(13)–13.5(12)+22.5(1) =10.5 ft
st=0 s =1.5(03)–13.5(02)+22.5(0) =0
t=0s
t=1s and
t=5s
4.50t2–27.0t+22.5 =0
v=ds
dt =4.50t2–27.0t+22.5
s|t=6s=1.5(63)–13.5(62)+22.5(6) =-27.0 ft
T
h
e pos
i
t
i
on of a part
i
c
l
e a
l
ong a stra
i
g
h
t
li
ne
i
s g
i
ven
b
y
, where tis in seconds.
Determine the position of the particle when and the
total distance it travels during the 6-s time interval. Hint:
Plot the path to determine the total distance traveled.
t=6s
s=(1.5t3–13.5t2+22.5t)ft
7
12–7.
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
dt
=2t–6
a=
dv
dt
=2
v=0
when
t=3
st=0=5
st=3=–4
st=6=5
vavg =
∆s
∆t
=
0
6
=0 Ans.
(
vsp
)
avg =
s
T
∆t
=
9+9
6
=3 m
>
s Ans.
a
t=6=2 m>s2
Ans.
Ans:
v
avg
=0
(v
sp
)
avg
=3 m>s
a
t=6 s =
2 m
>
s
2
*12–8.
A particle is moving along a straight line such that its
position is defined by , where tis in
seconds. Determine (a) the displacement of the particle
during the time interval from to , (b) the
average velocity of the particle during this time interval,
and (c) the acceleration when .t=1s
t=5st=1s
s=(10t2+20) mm
SOLUTION
s=10t2+20
12–9.
The acceleration of a particle as it moves along a straight
line is given by a
=
(2t
–
1) m
>
s
2
, 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
Lv
2
dv=
Lt
0
(2t –1)dt
v=t2–t+2
dx =v dt
Ls
t
ds =
Lt
0
(t2–t+2)dt
s=
1
3
t3–
1
2
t2+2t+1
When
t=6 s
v=32 m>s
Ans.
s=67 m
12–10.
SOLUTION
a=5
A
3s1
3+s5
2
B
Aparticle moves along a straight line with an acceleration
of , where sis in meters.
Determine the particle’s velocity when , if it starts
from rest when .Use a numerical method to evaluate
the integral.
s=1m
s=2m
a=5>(3s1>3+s5>2)m>s2
12–11.
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
*12–12.
SOLUTION
Ans.
Ans.s=0.792 km =792 m
(120)2=702+2(6000)(s–0)
v2=v1
2+2ac(s–s1)
t=8.33(10–3)hr=30 s
120 =70 +6000(t)
v=v1+act
Traveling with an initial speed of a car accelerates
at along a straight road. How long will it take to
reach a speed of Also, through what distance
does the car travel during this time?
120 km>h?
6000 km>h2
70 km
>
h,
12–13.
SOLUTION
stopping distance can be obtained using Eq. 12–6 with and .
Ans.
For a drunk driver, the car moves a distance of before he
or she reacts and decelerates the car.The stopping distance can be obtained using
Eq. 12–6 with and .
Ans.d=616 ft
02=442+2(–2)(d–132)
A
:
+
B
v2=v2
0+2ac(s–s0)
v=0s0=d¿=132 ft
d¿=vt=44(3) =132 ft
d=517 ft
02=442+2(–2)(d–33.0)
A
:
+
B
v2=v2
0+2ac(s–s0)
v=0s0=d¿=33.0 ft
Tests reveal that a normal driver takes about before
he or she can react to a situation to avoid a collision. It takes
about 3 s for a driver having 0.1% alcohol in his system to
do the same. If such drivers are traveling on a straight road
at 30 mph (44 ) and their cars can decelerate at ,
determine the shortest stopping distance dfor each from
the moment they see the pedestrians. Moral: If you must
drink, please don’t drive!
2ft>s2
ft>s
0.75 s
d
v144 ft/s
12–14.
The position of a particle along a straight-line path is
defined by s = (t3 – 6t2 – 15t + 7) ft, where t is in seconds.
Determine the total distance traveled when t = 10 s. What
are the particle’s average velocity, average speed, and the
instantaneous velocity and acceleration at this time?
12–15.
SOLUTION
Ans.
Ans.s=1
k
B¢
2kt +
¢
1
n2
0
≤≤
1
2
–1
n0
R
s=
2a2kt +a1
n2
0bb1
2
2k3t
0
Ls
0
ds =Lt
0
dt
a2kt +a1
v2
0bb1
2
ds =ndt
n=a2kt +a1
n2
0bb–1
2
–1
21n–2–n–2
02=-kt
Ln
n0
n–3dn=Lt
0
–kdt
a=
dn
dt
=-kn3
A part
i
c
l
e
i
s mov
i
ng w
i
t
h
a ve
l
oc
i
ty of w
h
en an
d
If it is subjected to a deceleration of
where kis a constant, determine its velocity and position as
functions of time.
a=-kv3,t=0.
s=0v
0
*12–16.
A part
i
c
l
e
i
s mov
i
ng a
l
ong a stra
i
g
h
t
li
ne w
i
t
h
an
i
n
i
t
i
a
l
velocity of when it is subjected to a deceleration of
, where vis in . Determine how far it
travels before it stops. How much time does this take?
m>sa=(–1.5v1>2)m>s2
6m>s
SOLUTION
1 7
12–17.
SOLUTION
(1)
For A:
(2)
Require the moment of closest approach.
Worst case without collision would occur when .
At , from Eqs. (1) and (2):
Ans.d=16.9 ft
157.5 =d+140.625
60(0.75) +60(3.75 –0.75) –7.5(3.75 –0.75)2=d+60(3.75) –6(3.75)2
t=3.75 s
sA=sB
t=3.75 s
60 –12t=60 –15(t–0.75)
vA=vB
sA=60(0.75) +60(t–0.75) –1
2(15) (t–0.75)2,[t70.74]
(:
+)s=s0+v0t+1
2act2
vA=60 –15(t–0.75),
[t70.75]
(:
+)v=v0+act
sB=d+60t–1
2(12) t2
(:
+)s=s0+v0t+1
2act2
vB=60 –12 t
(:
Car Bis traveling a distance dahead of car A.Both cars are
traveling at when the driver of Bsuddenly applies the
brakes,causing his car to decelerate at .It takes the
driver of car A0.75 s to react (this is the normal reaction
time for drivers). When he applies his brakes,he decelerates
at .Determine the minimum distance dbe tween the
cars so as to avoid a collision.
15 ft>s2
12 ft>s2
60 ft>s
d
AB
Ans:
d=16.9 ft
12–18.
The acceleration of a rocket traveling upward is given by
where sis in meters. Determine the
time needed for the rocket to reach an altitude of
Initially, and when t=0.s=0v=0s=100 m.
a=16+0.02s2m>s2,
12–19.
A train starts from rest at station Aand accelerates at
for 60 s. Afterwards it travels with a constant
velocity for 15 min. It then decelerates at 1 until it is
brought to rest at station B. Determine the distance
between the stations.
m>s2
0.5 m>s2
*12–20.
The velocity of a particle traveling along a straight line is
,where is in seconds.If when
,determine the position of the particle when .
What is the total distance traveled during the time interval
to ? Also, what is the acceleration when ?t=2 st=4 st=0
t=4 st=0
s=4 fttv =(3t2–6t) ft>s
SOLUTION