40 Solutions Manual
Problem 2.11
The position of a car as a function of time t, with t>0and expressed in seconds, is
Er.t/ DŒ.5:98t2C0:139t30:0149t4/O{C.0:523t2C0:0122t 30:00131t4/O|çft:
Determine the velocity, speed, and acceleration of the car for tD15 s.
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Dynamics 2e 41
Problem 2.12
The position of a car as a function of time t, with t>0and expressed in seconds, is
Er.t/ DŒ12:3.tC1:54e0:65t /O{C2:17.tC1:54e0:65t /O|çm:
Find the difference between the average velocity over the time interval
0t2
s and the true velocity
computed at the midpoint of the interval, i.e., at
tD1
s. Repeat the calculation for the time interval
8
s
t10
s. Explain why the difference between the average velocity and the true velocity over the
time interval 0t2s is not equal to that over 8st10 s.
Solution
The velocity is obtained by taking the derivative of the position with respect to time. This gives
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permission of McGraw-Hill, is prohibited.
42 Solutions Manual
Problem 2.13
The position of a car as a function of time t, with t>0and expressed in seconds, is
Er.t/ DŒ.66t 120/O{C.1:2 C31:7t 8:71t2/O|çft:
If the speed limit is 55 mph, determine the time at which the car will exceed this limit.
Solution
In order to solve the problem we need to determine the speed of the car. So, we first determine the velocity of
the car and then we compute its magnitude.
The velocity is found by taking the time derivative of the position. This yields,
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 43
Problem 2.14
The position of a car as a function of time t, with t>0and expressed in seconds, is
Er.t/ DŒ.66t 120/O{C.1:2 C31:7t 8:71t2/O|çft:
Determine the slope
✓
of the trajectory of the car for
t1D1
s and
t2D3
s. In addition, find the angle
between velocity and acceleration for
t1D1
s and
t2D3
s. Based on the values of
at
t1
and
t2
, argue
whether the speed of the car is increasing or decreasing at t1and t2.
Solution
Since the velocity is always tangent to the path, the angle
✓
can be computed by finding the velocity and
then determining the orientation of the velocity relative to the horizontal direction. The velocity is the time
derivative of the position. Differentiating the given expression for the position with respect to time, we have
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44 Solutions Manual
Problem 2.15
Let
ErDŒt O{C.2 C3t C2t2/O|çm
describe the motion of the point
P
relative to the Cartesian frame
of reference shown. Determine an analytic expression of the type
yDy.x/
for the trajectory of
P
for
0t5s.
Solution
The position of Pis given as
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Dynamics 2e 45
Problem 2.16
Let
ErDŒt O{C.2 C3t C2t2/O|çft
describe the motion of a point
P
relative to the Cartesian frame of
reference shown. Recalling that for any two vectors
Ep
and
Eq
we have that
EpEqDjEpjjEqjcos ˇ
, where
ˇ
is the angle formed by
Ep
and
Eq
, and recalling that the velocity vector is always tangent to the trajectory,
determine the function
.x/
describing the angle between the acceleration vector and the tangent to the
path of P.
Solution
The velocity vector is the time derivative of the position vector:
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46 Solutions Manual
Problem 2.17
The motion of a point
P
with respect to a Cartesian coordinate system is described by
ErD⇥2ptO{C
4ln.t C1/ C2t2O|⇤ft, where tdenotes time, t>0, and is expressed in seconds.
Determine the angle ✓formed by the tangent to the path and the horizontal direction at tD3s.
Solution
Since the velocity is always tangent to the path, we can find the angle
✓
by determining the angle formed by
the velocity vector and the horizontal direction. The velocity is the time derivative of the position. Hence,
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 47
Problem 2.18
The motion of a point
P
with respect to a Cartesian coordinate system is described by
ErD⇥2ptO{C
4ln.t C1/ C2t2O|⇤ft, where tdenotes time, t>0, and is expressed in seconds.
Determine the average acceleration of
P
between times
t1D4
s and
t2D6
s and find the difference
between it and the true acceleration of Pat tD5s.
Solution
By definition, the average acceleration over a time interval t1tt2is
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48 Solutions Manual
Problem 2.19
The motion of a stone thrown into a pond is described by
Er.t/ D⇥1:5 0:3e13:6t O{C0:094e13:6t 0:094 0:72t O|⇤m;
where
t
is time expressed in seconds, and
tD0
s is the time when the stone first hits the water. Determine
the stone’s velocity and acceleration. In addition, find the initial angle of impact
✓
of the stone with the
water, i.e., the angle formed by the stone’s trajectory and the horizontal direction at tD0.
Solution
The velocity of the stone is found by differentiating the position vector with respect to time. This gives
Dynamics 2e 49
Problem 2.20
As part of a mechanism, a peg
P
is made to slide within a rectilinear guide with the following prescribed
motion:
Er.t/ Dx0⇥sin.2⇡!t/ 3sin.⇡!t/⇤O{;
where
t
denotes time in seconds,
x0D1:2 in:
, and
!D0:5 rad=s
. Determine the displacement and the
distance traveled over the time interval
0t4
s. In addition, determine the corresponding average
velocity and average speed. Express displacement and distance traveled in
ft
, and express velocity and
speed in ft=s. You may find useful the following trigonometric identity: cos.2ˇ/ D2cos2ˇ1.
Solution
The function that describes Er.t / is the sum of two periodic functions. The period of the function sin.2⇡!t/
is half the period of the function
sin.⇡!t /
. Hence, the overall period
p
of
Er.t/
coincides with the period of
sin.⇡!t /. We determine pas follows:
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