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
A:
s = s0 + v0t +
s1 = 0 + 0 +
+ B
A:
1 2
at
2 c
1
(0.5)(602) = 900 m
2
v = v0 + act
laws
or
v1 = 0 + 0.5(60) = 30 m>s
in
teaching
Web)
For stage (2) motion, v0 = 30 m>s, s0 = 900 m, ac = 0 and t = 15(60) = 900 s. Thus,
1 2
at
2 c
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permitted.
Dissemination
s = s0 + v0t +
copyright
+ B
A:
use
United
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learning.
is
of
the
not
s2 = 900 + 30(900) + 0 = 27 900 m
the
v = v0 + act
assessing
is
work
solely
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protected
+ B
A:
(including
for
work
student
the
and
n ac = - 1 m>s2. Thus,
For stage (3) motion, v0 = 30 m>s, v = 0, s0 = 27 900 m and
provided
integrity
of
and
work
this
0 = 30 + (- 1)t
destroy
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and
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1 2
at
2 c
their
s = s0 + v0t +
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:
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courses
part
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t = 30 s
s3 = 27 900 + 30(30) +
= 28 350 m = 28.4 km
1
(- 1)(302)
2
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
v =
ds
= 3t2 - 18t + 15
dt
v = 0 when t = 1 s and t = 5 s
t = 0, s = 0
t = 1 s, s = 7 ft
Ans.
sT = 7 + 7 + 25 + (25 - 18) = 46 ft
Ans.
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is
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by
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use
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learning.
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permitted.
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laws
t = 6 s, s = - 18 ft
or
t = 5 s, s = - 25 ft
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
A
45
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
(a ) t2
2 cy
1
d sin 10° = 0 + 65.53t + (-32.2)t2
2
laws
or
sy = (s0)y + (v0)y t +
in
Web)
(2)
copyright
(+ c)
permitted.
Dissemination
Wide
Solving Eqs. (1) and (2) yields
Ans.
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States
d = 166 ft
sale
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the
(including
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the
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on
learning.
is
t = 3.568 s
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,
where t is in seconds. If he starts from rest at A, determine
the magnitudes of his velocity and acceleration when he
reaches B.
300 ft
A
60°
300 ft
SOLUTION
v
L0
B
t
dv =
L0
0.3tdt
v = 0.15t2
s
L0
t
ds =
L0
0.15t2 dt
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
or
Ans.
in
teaching
Web)
laws
#
at = v = 0.3t|t = 18.453 s = 5.536 ft>s2
51.082
v2
=
= 8.696 ft>s2
r
300
States
World
permitted.
Dissemination
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copyright
an =
destroy
of
and
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courses
part
the
is
This
provided
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this
assessing
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the
(including
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work
student
the
by
and
use
United
on
learning.
is
of
the
Ans.
will
not
instructors
(5.536)2 + (8.696)2 = 10.3 ft s2
their
a2t + a2n =
sale
a =
Q5
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
Web)
laws
or
#
Here, at = vA = 3 m>s. Thus, the magnitude of acceleration is
teaching
32 + 0.52472 = 3.05 m s2
this
assessing
is
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the
(including
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student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
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copyright
in
Ans.
integrity
of
and
work
provided
any
courses
part
the
is
This
destroy
of
and
will
their
a2t + a2n =
sale
a =
sale
will
their
destroy
of
and
any
courses
part
the
is
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
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student
the
by
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or
Q6
Determine the displacement of the block B if A is pulled
down 4 ft.
C
SOLUTION
A
2 sA + 2 sC = l1
¢sA = - ¢sC
B
sB - sC + sB = l2
2 ¢sB = ¢sC
Thus,
2 ¢sB = - ¢sA
2 ¢sB = - 4
¢sB = - 2 ft = 2 ft c
Ans.
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
A
B
0.3 m/s
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.
4
ds B ds A
+
+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.
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
speed of 105 ft>s and is decreasing its speed at 25 ft>s2.
Determine the relative velocity and relative acceleration of
car A with respect to car B at this instant.
vA
A
B
vB
rA
300 ft
60
rB
SOLUTION
vA = vB + vA>B
- 90i = -105 sin 30° i + 105 cos 30°j + 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
b = 67.6° d
37.5
Ans.
u = tan - 1 a
aA = aB + aA>B
300
or
j = 25 cos 60°i - 25 sin 60°j - 44.1 sin 60°i - 44.1 cos 60°j + aA>B
teaching
Web)
- 15i -
19022
Wide
copyright
in
aA>B = {10.69i + 16.70j} ft>s2
aA>B = 2(10.69)2 + (16.70)2 = 19.8 ft>s2
instructors
States
World
permitted.
Ans.
of
the
not
16.70
b = 57.4° a
10.69
on
learning.
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United
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student
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the
(including
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is
This
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u = tan - 1 a
250 ft
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
30
SOLUTION
Relative Velocity:
vB = vA + vB>A
50 sin 30°i + 50 cos 30°j = 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.
laws
or
The direction of the relative velocity is the same as the direction of that for relative
acceleration. Thus
26.70
= 46.9° c
25.0
in
teaching
Web)
Ans.
is
of
the
not
World
Relative Acceleration: Since car B is traveling along a curve, its normal
y2B
502
acceleration is (aB)n =
= 3571.43 mi>h2. Applying Eq. 12–35 gives
=
r
0.7
the
by
and
use
United
on
learning.
aB = aA + aB>A
the
(including
work
(1100 sin 30° + 3571.43 cos 30°)i + (1100 cos 30° - 3571.43 sin 30°)j = 0 + aB>A
assessing
is
work
of
aB>A = {3642.95i - 833.09j} mi>h2
provided
integrity
of
and
this
Thus, the magnitude of the relative velocity aB/A is
Ans.
and
any
courses
the
is
aB>A = 23642.952 + (- 833.09)2 = 3737 mi>h2
their
destroy
of
And its direction is
will
833.09
= 12.9° c
3642.95
sale
f = tan - 1
Ans.
permitted.
Wide
copyright
u = tan - 1
B
50 mi/h