Q1

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 1 2
A: s = s0 + v0t + at
2 c

1
s1 = 0 + 0 + (0.5)(602) = 900 m
2

+ B
A: v = v0 + act

v1 = 0 + 0.5(60) = 30 m>s

or
laws
For stage (2) motion, v0 = 30 m>s, s0 = 900 m, ac = 0 and t = 15(60) = 900 s. Thus,

Web)
teaching
in

+ B 1 2
A: s = s0 + v0t +
copyright

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at
2 c
Dissemination

permitted.
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States

instructors

s2 = 900 + 30(900) + 0 = 27 900 m

not
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of

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learning.

on
United

use

For stage (3) motion, v0 = 30 m>s, v = 0, s0 = 27 900 m and

n ac = - 1 m>s2. Thus,
and
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+ B
A:
(including

v = v0 + act
the
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assessing

0 = 30 + (- 1)t
this
work

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integrity
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t = 30 s
This

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1 2
any
and

+ s = s0 + v0t + at
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their

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1
s3 = 27 900 + 30(30) + (- 1)(302)
2

= 28 350 m = 28.4 km Ans.

Q2

The position of a particle on a straight line is given by

s = 1t3 - 9t2 + 15t2 ft, where t is in seconds. Determine the
position of the particle when t = 6 s and the total distance it
travels during the 6-s time interval. Hint: Plot the path to
determine the total distance traveled.

SOLUTION
s = t3 - 9t2 + 15t

ds
v = = 3t2 - 18t + 15
dt

v = 0 when t = 1 s and t = 5 s

t = 0, s = 0

t = 1 s, s = 7 ft

t = 5 s, s = - 25 ft

t = 6 s, s = - 18 ft Ans.

or
laws
sT = 7 + 7 + 25 + (25 - 18) = 46 ft Ans.

Web)
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copyright

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Dissemination

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Q3

A golf ball is struck with a velocity of 80 ft>s as shown.

Determine the distance d to where it will land.

vA 80 ft/s
B
45
A 10
d
SOLUTION
Horizontal Motion: The horizontal component of velocity is (v0)x = 80 cos 55
= 45.89 ft>s.The initial and final horizontal positions are (s0)x = 0 and sx = d cos 10,
respectively.
+ B
A: sx = (s0)x + (v0)x t

d cos 10 = 0 + 45.89t (1)

Vertical Motion: The vertical component of initial velocity is (v0)y = 80 sin 55

= 65.53 ft>s. The initial and final vertical positions are (s0)y = 0 and sy = d sin 10,
respectively.
1
(+ c) sy = (s0)y + (v0)y t + (a ) t2
2 cy

or
laws
1
d sin 10 = 0 + 65.53t + (-32.2)t2 (2)

Web)
2
in
copyright

Solving Eqs. (1) and (2) yields

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Dissemination

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d = 166 ft Ans.
States

instructors

not
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t = 3.568 s
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learning.

on
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Q4

When the motorcyclist is at A, he increases his speed along

#
the vertical circular path at the rate of v = 10.3t2 ft>s2,
300 ft 60
where t is in seconds. If he starts from rest at A, determine
the magnitudes of his velocity and acceleration when he
reaches B. A
300 ft

SOLUTION
v t B
dv = 0.3tdt
L0 L0

v = 0.15t2
s t
ds = 0.15t2 dt
L0 L0

s = 0.05t3

When s = p3 (300) ft, p

3 (300) = 0.05t3 t = 18.453 s

v = 0.15(18.453)2 = 51.08 ft>s = 51.1 ft>s Ans.

or
#

laws
at = v = 0.3t|t = 18.453 s = 5.536 ft>s2

Web)
teaching
in

v2 51.082
an = = = 8.696 ft>s2
copyright

Wide
r 300
Dissemination

permitted.
World
States

a = a2t + a2n = (5.536)2 + (8.696)2 = 10.3 ft s2

instructors

Ans.
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Q5

A toboggan is traveling down along a curve which can be y

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
# y = 0.01x2
vA = 3 m>s2.

A
36 m

SOLUTION x
60 m
Acceleration: The radius of curvature of the path at point A must be determined
dy d2y
first. Here, = 0.02x and 2 = 0.02, then
dx dx

[1 + (dy>dx)2]3>2 [1 + (0.02x)2]3>2
r = = 2 = 190.57 m
|d2y>dx2| |0.02| x = 60 m

To determine the normal acceleration, apply Eq. 1220.

v2 102
an = = = 0.5247 m>s2
r 190.57
#
Here, at = vA = 3 m>s. Thus, the magnitude of acceleration is

or
laws

Web)
a = a2t + a2n = 32 + 0.52472 = 3.05 m s2
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Ans.
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 Q6

Thus,
down 4 ft.

2 sB = - 4
2 sB = sC
sA = - sC

2 sB = - sA
2 sA + 2 sC = l1
SOLUTION

sB - sC + sB = l2

sB = - 2 ft = 2 ft c
and

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Determine the displacement of the block B if A is pulled

this

by

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the

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of

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and

Dissemination

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Ans.

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Web)

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A
C

B
Q-7At what rate, and in which direction must weight A move if weight B is to fall at a rate of 0.3
m/s?

DATUM

sB

sA

0.3 m/s
B
A

Solution
There is a single cord between weights A and B, so the motion of one of the weights is dependent
upon one the other. While the physical system is somewhat more complex than the original, 2-pulley
example, the solution is not very different. The total length of the cord is made up of 9 segments, as
shown in the figure below. There are 4 equivalent-length sections that lengthen as weight B is
lowered. The length of each of these sections is labeled SB.

The length of the cord segment connected to weight A is labeled SA. There are four cord segments
around the four pulleys that do not change as the weights move. These segments are shown in red in
the figure, and their lengths are combined and called Lconst. The total cord length, then, is
4 s B + s A + L const = L total

Differentiating this equation with respect to time yields the time rates of change of position, or
velocities of each weight.
ds B ds A
4 + +0=0
dt dt
4 v B = v A

So, if weight B moves down at 0.3 m/s, weight A will move up at 1.2 m/s.
Q8

Cars A and B are traveling around the circular race track. vA

At the instant shown, A has a speed of 90 ft>s and is
increasing its speed at the rate of 15 ft>s2, whereas B has a A
speed of 105 ft>s and is decreasing its speed at 25 ft>s2. B
Determine the relative velocity and relative acceleration of vB
car A with respect to car B at this instant. rA 300 ft 60

rB 250 ft

SOLUTION
vA = vB + vA>B

- 90i = -105 sin 30 i + 105 cos 30j + vA>B

vA>B = 5-37.5i - 90.93j6 ft>s

vA/B = 2( -37.5)2 + ( -90.93)2 = 98.4 ft>s Ans.

90.93
u = tan - 1 a b = 67.6 d Ans.
37.5

aA = aB + aA>B

19022
- 15i - j = 25 cos 60i - 25 sin 60j - 44.1 sin 60i - 44.1 cos 60j + aA>B

or
300

Web)
teaching
aA>B = {10.69i + 16.70j} ft>s2
in
copyright

Wide
aA>B = 2(10.69)2 + (16.70)2 = 19.8 ft>s2 Ans.
permitted.
World
States

instructors

16.70
not

u = tan - 1 a b = 57.4 a
the
of

Ans.
learning.

10.69
on
United

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Q9

At the instant shown, cars A and B travel at speeds of

70 mi>h and 50 mi>h, respectively. If B is increasing its
speed by 1100 mi>h2, while A maintains a constant speed,
determine the velocity and acceleration of B with respect to
A.Car B moves along a curve having a radius of curvature of
0.7 mi. A
vA 70 mi/h
vB 50 mi/h

B
SOLUTION 30

Relative Velocity:
vB = vA + vB>A

50 sin 30i + 50 cos 30j = 70j + vB>A

vB>A = {25.0i - 26.70j} mi>h

Thus, the magnitude of the relative velocity vB/A is

yB>A = 225.02 + ( - 26.70)2 = 36.6 mi>h Ans.

The direction of the relative velocity is the same as the direction of that for relative
acceleration. Thus

or
laws
26.70
u = tan - 1 = 46.9 c Ans.
25.0

Web)
teaching
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copyright

Wide
Relative Acceleration: Since car B is traveling along a curve, its normal
y2B 502 permitted.

acceleration is (aB)n = = = 3571.43 mi>h2. Applying Eq. 1235 gives

World

r 0.7
not
the
of

is

aB = aA + aB>A
learning.

on
United

use

and
by

the

(1100 sin 30 + 3571.43 cos 30)i + (1100 cos 30 - 3571.43 sin 30)j = 0 + aB>A
work
(including

the

aB>A = {3642.95i - 833.09j} mi>h2

of
work
is

assessing

Thus, the magnitude of the relative velocity aB/A is

this
and

of

integrity
provided

aB>A = 23642.952 + (- 833.09)2 = 3737 mi>h2 Ans.

is

the
courses

any
and

And its direction is

of

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833.09
f = tan - 1
will

= 12.9 c Ans.
sale

3642.95