1. The position of a particle in milimeters is given by s=27−12t+t2, where t
is in seconds. Plot the s−t and v−t relationships for the first 9 seconds.
Determine the net displacement ∆s during that interval and the total
distance D traveled.
2. Ball 1 is launched with an initial vertical velocity v1=50 m/s. Three seconds later,
ball 2 is launched with an initial vertical velocity v2. Determine v2 if the balls are
collide at an altitude of 90 m. At the instant of collision, is ball 1 ascending or
descending?
3. A motorcycle patrolman starts from rest at A two seconds after a car, speeding
at the constant rate of 120 km/h, passes point A. If the patrolman accelerates at the
rate of 6 m/s2 until he reaches his maximum permissible speed of 150 km/h, which
he maintains, calculate the distance s from point A to the point at which he
overtakes the car.
4. The brake mechanism shown in the figure is composed of a piston moving in a
fixed cylinder filled with oil. When the brake pedal is pressed while the vehicle
moves with a speed v0, the piston moves, oil passes through the channels inside the
piston and the vehicle slows down in proportion to its speed, a=−kv. Determine a) v
in terms of t, b) x in terms of t, c) v in terms of x. Also construct the related graphics.
Piston
Oil
5. A bumper, consisting of a nest of three
springs, is used to arrest the horizontal
motion of a large mass which is traveling at
40 m/s as it contacts the bumper. The two
outer
springs
cause
a
deceleration
proportional to the spring deformation. The
center spring increases the deceleration rate
when the compression exceeds 0.5 m as
shown
on
the
graph.
Determine
the
maximum compression x of the outer
springs.
6. A ball is thrown vertically upward with an initial speed of
25 m/s from the base A of a 15-m cliff. Determine the distance
h by which the ball clears the top of the cliff and the time t
after release for the ball to land at B. Also, calculate the impact
velocity vB. Neglect air resistance and the small horizontal
motion of the ball.
C
At A vA=25 m/s
g: gravitational acceleration
y
At C (at apex) vC=0
g=9.81 m/s2
Acceleration : constant.
A
A−C :
vC2 = v 2A + 2a( y − y o ) ⇒
h = 16.86 m
0 = 25 2 + 2(− 9.81)(15 + h − 0 )
A−B : y B = 15 m
y B = y A + v At +
15 = 0 + 25(t ) +
t1 = 4.4 s


1 2
at
2
1
(− 9.81)t 2 ⇒
2
,
t 2 = 0.69 s
4.905t 2 − 25t + 15 = 0
(represents the second time at which y=15 m)
v B = v A + at ⇒
v B = 25 − 9.81(4.4 ) = −18.16 m / s (downward
)
7. The body falling with speed strikes vo and maintains contact with the platform
supported by a nest of springs. The acceleration of the body after impact is a=g – cy
where c is a positive constant and y is measured from the original platform position.
If the maximum compression of the springs is observed to be ym, determine the
constant c.
Solution:
a = g − cy
0
ym
vo
0
vdv = ady
∫ vdv = ∫ (g − cy )dy
0
⇒
ym
1 2
1

v  = gy − cy 2 
2  vo
2
0
1 2
1
2
0 − v0 = gym − cym
2
2
1
1 2
2
cym = v0 + gym
⇒
2
2
2
v0 + 2 gym
c=
2
ym
8. A test projectile is fired horizontally into a viscous liquid with a velocity vo.
The retarding force is proportional to the square of the velocity, so that the
acceleration becomes a=−kv2. Derive expressions for the distance D traveled in
the liquid and the corresponding time t required to reduce the velocity to vo/2.
Neglect any vertical motion.
Solution:
2
a = −kv
v
∫
vdv
v o − kv
2
x
= ∫ dx
0
vdv = adx
a = f (v ) ( given )
⇒
⇒
1
x = − lnv ]vv ⇒
o
k
x=
vdv
= dx
a
1 vo
ln
k
v
Derive expressions for the distance D traveled in the liquid and the corresponding
time t required to reduce the velocity to vo/2.
x=
1 vo
ln
k v
when v =
x=
vo
2
1 vo
ln
k v
1 1 kx
e
vo k
for x = D
t=
x=D=
⇒
dx
⇒
v=
dt
t=
0.693
1 vo 1
= ln 2 =
ln
k vo k
k
2
v
− kx = ln
⇒
v = vo e − kx
vo
⇒
dx
dt =
v
]
x
0
=
(
⇒
dx
dt =
vo e − kx
{v = f (x )}
t
⇒
x
dx
− kx
v
e
0 o
∫ dt = ∫
0
)
1 kx
e −1
kvo
⇒
1
(2 − 1) = 1
kvo
kvo
1
x = ln 2
k
⇒
kx = ln 2
⇒
e kx = 2