Problem 13.1 In Example 13.2, suppose that the vehi-
cle is dropped from a height h = 6m. (a) What is the
downward velocity 1 s after it is released? (b) What is
its downward velocity just before it reaches the ground?
Solution: The equations that govern the motion are:
a = −g = −9.81 m/s2
v = −gt
s = − 21 gt 2 + h h
(a) v = −gt = −(9.81 m/s2 )(1 s) = −9.81 m/s.
The downward velocity is 9.81 m/s.
(b) We need to first determine the time at which the vehicle hits the
ground
2h 2(6 m)
s = 0 = − 12 gt 2 + h ⇒ t = = = 1.106 s
g 9.81 m/s2
Now we can solve for the velocity
v = −gt = −(9.81 m/s2 )(1.106 s) = −10.8 m/s.
The downward velocity is 10.8 m/s.
Problem 13.2 The milling machine is programmed so
that during the interval of time from t = 0 to t = 2 s,
the position of its head (in inches) is given as a function
of time by s = 4t − 2t 3 . What are the velocity (in in/s)
and acceleration (in in/s2 ) of the head at t = 1 s?
Solution: The motion is governed by the equations
s = (4 in/s)t − (2 in/s2 )t 2 , s
v = (4 in/s) − 2(2 in/s2 )t,
a = −2(2 in/s2 ).
At t = 1 s, we have v = 0, a = −4 in/s2 .
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, Problem 13.3 In an experiment to estimate the accel-
eration due to gravity, a student drops a ball at a distance
s
of 1 m above the floor. His lab partner measures the time
it takes to fall and obtains an estimate of 0.46 s.
(a) What do they estimate the acceleration due to grav-
ity to be?
(b) Let s be the ball’s position relative to the floor.
Using the value of the acceleration due to gravity
that they obtained, and assuming that the ball is
released at t = 0, determine s (in m) as a function
of time.
Solution: The governing equations are
a = −g
v = −gt
s = − 12 gt 2 + h s0
(a) When the ball hits the floor we have
2h 2(1 m)
0 = − 12 gt 2 + h ⇒ g = = = 9.45 m/s2
t2 (0.46 s)2
g = 9.45 m/s2
(b) The distance s is then given by
s = − 12 (9.45 m/s2 ) + 1 m. s = −(4.73 m/s2 )t 2 + 1.0 m.
Problem 13.4 The boat’s position during the interval
of time from t = 2 s to t = 10 s is given by s = 4t +
1.6t 2 − 0.08t 3 m.
(a) Determine the boat’s velocity and acceleration at
t = 4 s.
(b) What is the boat’s maximum velocity during this
interval of time, and when does it occur?
Solution:
s = 4t + 1.6t 2 − 0.08t 3
a) v(4s) = 12.96 m/s2
ds a(4s) = 1.28 m/s2
v= = 4 + 3.2t − 0.24t 2 ⇒
dt b) a = 3.2 − 0.48t = 0 ⇒ t = 6.67s
v(6.67s) = 14.67 m/s
dv
a= = 3.2 − 0.48t
dt
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, Problem 13.5 The rocket starts from rest at t = 0 and
travels straight up. Its height above the ground as a
function of time can be approximated by s = bt 2 + ct 3 ,
where b and c are constants. At t = 10 s, the rocket’s
velocity and acceleration are v = 229 m/s and a = 28.2
m/s2 . Determine the time at which the rocket reaches
supersonic speed (325 m/s). What is its altitude when
that occurs?
Solution: The governing equations are
s = bt 2 + ct 3 ,
s
v = 2bt + 3ct 2 ,
a = 2b + 6ct.
Using the information that we have allows us to solve for the constants
b and c.
(229 m/s) = 2b(10 s) + 3c(10 s)2 ,
(28.2 m/s2 ) = 2b + 6c(10 s).
Solving these two equations, we find b = 8.80 m/s2 , c = 0.177 m/s3 .
When the rocket hits supersonic speed we have
(325 m/s) = 2(8.80 m/s2 )t + 3(0.177 m/s3 )t 2 ⇒ t = 13.2 s.
The altitude at this time is
s = (8.80 m/s2 )(13.2 s)2 + (0.177 m/s3 )(13.2 s)3 s = 1940 m.
Problem 13.6 The position of a point during the inter-
val of time from t = 0 to t = 6 s is given by s = − 12 t 3 +
6t 2 + 4t m.
(a) What is the maximum velocity during this interval
of time, and at what time does it occur?
(b) What is the acceleration when the velocity is a
maximum?
dv
Solution: Maximum velocity occurs where a = = 0 (it could be a minimum)
dt
da
This occurs at t = 4 s. At this point = −3 so we have a maximum.
s = − 21 t 3 + 6t 2 + 4t m dt
(a) Max velocity is at t = 4 s. where v = 28 m/s and
v = − 23 t 2 + 12t + 4 m/s
(b) a = 0 m/s2
a = −3t + 12 m/s2
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, Problem 13.7 The position of a point during the inter-
val of time from t = 0 to t = 3 seconds is s = 12 +
5t 2 − t 3 m.
(a) What is the maximum velocity during this interval
of time, and at what time does it occur?
(b) What is the acceleration when the velocity is a
maximum?
Solution:
d2v
ds This is indeed a maximum, since = −6 < 0. The maximum
(a) The velocity is = 10t − 3t 2 . The maximum occurs when dt 2
dt velocity is
dv
= 10 − 6t = 0, from which
dt v = 10t − 3t 2 t=1.667 = 8.33 m /s
10
t= = 1.667 seconds.
6 dv
(b) The acceleration is = 0 when the velocity is a maximum.
dt
Problem 13.8 The rotating crank causes the position
of point P as a function of time to be s = 0.4 sin
(2π t) m. P
(a) Determine the velocity and acceleration of P at
t = 0.375 s. s
(b) What is the maximum magnitude of the velocity
of P ?
(c) When the magnitude of the velocity of P is a
maximum, what is the acceleration of P ?
Solution:
s = 0.4 sin(2π t)
a) v(0.375s) = −1.777 m/s
ds a(0.375) = −11.2 m/s2
v= = 0.8π cos(2π t) ⇒
dt b) vmax = 0.8π = 2.513 m/s2
c) vmax ⇒ t = 0, nπ ⇒ a = 0
dv
a= = −1.6π 2 sin(2π t)
dt
Problem 13.9 For the mechanism in Problem 13.8, Solution:
draw graphs of the position s, velocity v, and acce-
leration a of point P as functions of time for 0 ≤ t ≤ 2 s.
Using your graphs, confirm that the slope of the graph
of s is zero at times for which v is zero, and the slope
of the graph of v is zero at times for which a is zero.
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cle is dropped from a height h = 6m. (a) What is the
downward velocity 1 s after it is released? (b) What is
its downward velocity just before it reaches the ground?
Solution: The equations that govern the motion are:
a = −g = −9.81 m/s2
v = −gt
s = − 21 gt 2 + h h
(a) v = −gt = −(9.81 m/s2 )(1 s) = −9.81 m/s.
The downward velocity is 9.81 m/s.
(b) We need to first determine the time at which the vehicle hits the
ground
2h 2(6 m)
s = 0 = − 12 gt 2 + h ⇒ t = = = 1.106 s
g 9.81 m/s2
Now we can solve for the velocity
v = −gt = −(9.81 m/s2 )(1.106 s) = −10.8 m/s.
The downward velocity is 10.8 m/s.
Problem 13.2 The milling machine is programmed so
that during the interval of time from t = 0 to t = 2 s,
the position of its head (in inches) is given as a function
of time by s = 4t − 2t 3 . What are the velocity (in in/s)
and acceleration (in in/s2 ) of the head at t = 1 s?
Solution: The motion is governed by the equations
s = (4 in/s)t − (2 in/s2 )t 2 , s
v = (4 in/s) − 2(2 in/s2 )t,
a = −2(2 in/s2 ).
At t = 1 s, we have v = 0, a = −4 in/s2 .
c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.
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, Problem 13.3 In an experiment to estimate the accel-
eration due to gravity, a student drops a ball at a distance
s
of 1 m above the floor. His lab partner measures the time
it takes to fall and obtains an estimate of 0.46 s.
(a) What do they estimate the acceleration due to grav-
ity to be?
(b) Let s be the ball’s position relative to the floor.
Using the value of the acceleration due to gravity
that they obtained, and assuming that the ball is
released at t = 0, determine s (in m) as a function
of time.
Solution: The governing equations are
a = −g
v = −gt
s = − 12 gt 2 + h s0
(a) When the ball hits the floor we have
2h 2(1 m)
0 = − 12 gt 2 + h ⇒ g = = = 9.45 m/s2
t2 (0.46 s)2
g = 9.45 m/s2
(b) The distance s is then given by
s = − 12 (9.45 m/s2 ) + 1 m. s = −(4.73 m/s2 )t 2 + 1.0 m.
Problem 13.4 The boat’s position during the interval
of time from t = 2 s to t = 10 s is given by s = 4t +
1.6t 2 − 0.08t 3 m.
(a) Determine the boat’s velocity and acceleration at
t = 4 s.
(b) What is the boat’s maximum velocity during this
interval of time, and when does it occur?
Solution:
s = 4t + 1.6t 2 − 0.08t 3
a) v(4s) = 12.96 m/s2
ds a(4s) = 1.28 m/s2
v= = 4 + 3.2t − 0.24t 2 ⇒
dt b) a = 3.2 − 0.48t = 0 ⇒ t = 6.67s
v(6.67s) = 14.67 m/s
dv
a= = 3.2 − 0.48t
dt
c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.
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, Problem 13.5 The rocket starts from rest at t = 0 and
travels straight up. Its height above the ground as a
function of time can be approximated by s = bt 2 + ct 3 ,
where b and c are constants. At t = 10 s, the rocket’s
velocity and acceleration are v = 229 m/s and a = 28.2
m/s2 . Determine the time at which the rocket reaches
supersonic speed (325 m/s). What is its altitude when
that occurs?
Solution: The governing equations are
s = bt 2 + ct 3 ,
s
v = 2bt + 3ct 2 ,
a = 2b + 6ct.
Using the information that we have allows us to solve for the constants
b and c.
(229 m/s) = 2b(10 s) + 3c(10 s)2 ,
(28.2 m/s2 ) = 2b + 6c(10 s).
Solving these two equations, we find b = 8.80 m/s2 , c = 0.177 m/s3 .
When the rocket hits supersonic speed we have
(325 m/s) = 2(8.80 m/s2 )t + 3(0.177 m/s3 )t 2 ⇒ t = 13.2 s.
The altitude at this time is
s = (8.80 m/s2 )(13.2 s)2 + (0.177 m/s3 )(13.2 s)3 s = 1940 m.
Problem 13.6 The position of a point during the inter-
val of time from t = 0 to t = 6 s is given by s = − 12 t 3 +
6t 2 + 4t m.
(a) What is the maximum velocity during this interval
of time, and at what time does it occur?
(b) What is the acceleration when the velocity is a
maximum?
dv
Solution: Maximum velocity occurs where a = = 0 (it could be a minimum)
dt
da
This occurs at t = 4 s. At this point = −3 so we have a maximum.
s = − 21 t 3 + 6t 2 + 4t m dt
(a) Max velocity is at t = 4 s. where v = 28 m/s and
v = − 23 t 2 + 12t + 4 m/s
(b) a = 0 m/s2
a = −3t + 12 m/s2
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, Problem 13.7 The position of a point during the inter-
val of time from t = 0 to t = 3 seconds is s = 12 +
5t 2 − t 3 m.
(a) What is the maximum velocity during this interval
of time, and at what time does it occur?
(b) What is the acceleration when the velocity is a
maximum?
Solution:
d2v
ds This is indeed a maximum, since = −6 < 0. The maximum
(a) The velocity is = 10t − 3t 2 . The maximum occurs when dt 2
dt velocity is
dv
= 10 − 6t = 0, from which
dt v = 10t − 3t 2 t=1.667 = 8.33 m /s
10
t= = 1.667 seconds.
6 dv
(b) The acceleration is = 0 when the velocity is a maximum.
dt
Problem 13.8 The rotating crank causes the position
of point P as a function of time to be s = 0.4 sin
(2π t) m. P
(a) Determine the velocity and acceleration of P at
t = 0.375 s. s
(b) What is the maximum magnitude of the velocity
of P ?
(c) When the magnitude of the velocity of P is a
maximum, what is the acceleration of P ?
Solution:
s = 0.4 sin(2π t)
a) v(0.375s) = −1.777 m/s
ds a(0.375) = −11.2 m/s2
v= = 0.8π cos(2π t) ⇒
dt b) vmax = 0.8π = 2.513 m/s2
c) vmax ⇒ t = 0, nπ ⇒ a = 0
dv
a= = −1.6π 2 sin(2π t)
dt
Problem 13.9 For the mechanism in Problem 13.8, Solution:
draw graphs of the position s, velocity v, and acce-
leration a of point P as functions of time for 0 ≤ t ≤ 2 s.
Using your graphs, confirm that the slope of the graph
of s is zero at times for which v is zero, and the slope
of the graph of v is zero at times for which a is zero.
c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.
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