Mechanical Engineering Chapter 13 Problem Example Suppose That The Vehi Cle Dropped From Height What Thedownward

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

Problem 13.2 The milling machine is programmed so

that during the interval of time from t=0tot=2s,

the position of its head (in inches) is given as a function

of time by s=4t−2t3. What are the velocity (in in/s)

and acceleration (in in/s2) of the head at t=1s?

Solution: The motion is governed by the equations

Problem 13.3 In an experiment to estimate the accel-

eration due to gravity, a student drops a ball at a distance

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 sbe 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.

s

Solution: The governing equations are

(b) The distance sis then given by

1

Problem 13.4 The boat’s position during the interval

of time from t=2stot=10 s is given by s=4t+

1.6t2−0.08t3m.

(a) Determine the boat’s velocity and acceleration at

t=4s.

(b) What is the boat’s maximum velocity during this

interval of time, and when does it occur?

Solution:

10

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=bt2+ct3,

where band care 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=bt2+ct3,

Problem 13.6 The position of a point during the inter-

val of time from t=0tot=6 s is given by s=−

1

2t3+

6t2+4tm.

(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?

Problem 13.7 The position of a point during the inter-

val of time from t=0tot=3 seconds is s=12 +

5t2−t3ft.

(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?

Problem 13.8 The rotating crank causes the position

t=0.375 s.

(b) What is the maximum magnitude of the velocity

of P?

(c) When the magnitude of the velocity of Pis a

maximum, what is the acceleration of P?

s

v=ds

dt =0.8πcos(2πt) ⇒

a(0.375)=−11.2 m/s2

b) vmax =0.8π=2.513 m/s2

Problem 13.9 For the mechanism in Problem 13.8,

draw graphs of the position s, velocity v, and acce-

Solution:

12

c

currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Problem 13.10 A seismograph measures the horizon-

tal motion of the ground during an earthquake. An engi-

neer analyzing the data determines that for a 10-s interval

of time beginning at t=0, the position is approximated

by s=100 cos(2πt) mm. What are (a) the maximum

velocity and (b) maximum acceleration of the ground

during the 10-s interval?



ds

dt t=(2n−1)

4

=[0.2π]=0.628 m/s.

dt t=nπ

2

Problem 13.11 In an assembly operation, the robot’s

arm moves along a straight horizontal line. During an

interval of time from t=0tot=1 s, the position of the

arm is given by s=30t2−20t3mm. (a) Determine the

maximum velocity during this interval of time. (b) What

are the position and acceleration when the velocity is a

maximum?

s

Solution:

v=(60)1

Problem 13.12 In Active Example 13.1, the accelera-

tion (in m/s2) of point Prelative to point Ois given as

Solution: The governing equations are

a=(3 m/s4)t2

Problem 13.13 The Porsche starts from rest at time

t=0. During the ﬁrst 10 seconds of its motion, its

velocity in km/h is given as a function of time by v=

22.8t−0.88t2, where tis in seconds. (a) What is the

car’s maximum acceleration in m/s2, and when does it

occur? (b) What distance in km does the car travel dur-

ing the 10 seconds?

Solution: First convert the numbers into meters and seconds

14

Problem 13.14 The acceleration of a point is a=

20tm/s2. When t=0,s =40 m and v=−10 m/s.

What are the position and velocity at t=3s?

where C2is the constant of integration.

s=(10t2−10)dt +C2=10

3t3−10t+C2.

Problem 13.15 The acceleration of a point is a=

60t−36t2ft/s2. When t=0,s =0 and v=20 ft/s.

What are position and velocity as a function of time?

Problem 13.16 As a ﬁrst approximation, a bioengineer

studying the mechanics of bird ﬂight assumes that the

snow petrel takes off with constant acceleration. Video

measurements indicate that a bird requires a distance of

4.3 m to take off and is moving at 6.1 m/s when it does.

What is its acceleration?

Solution: The governing equations are

Problem 13.17 Progressively developing a more real-

istic model, the bioengineer next models the acceleration

of the snow petrel by an equation of the form a=

C(1+sin ωt), where Cand ωare constants. From video

measurements of a bird taking off, he estimates that

ω=18/s and determines that the bird requires 1.42 s

to take off and is moving at 6.1 m/s when it does. What

is the constant C?

16

Problem 13.18 Missiles designed for defense against

ballistic missiles have attained accelerations in excess

of 100 g’s, or 100 times the acceleration due to gravity.

Suppose that the missile shown lifts off from the ground

and has a constant acceleration of 100 g’s. How long

does it take to reach an altitude of 3000 m? How fast is

it going when it ranches that altitude?

Problem 13.19 Suppose that the missile shown lifts

off from the ground and, because it becomes lighter as

its fuel is expended, its acceleration (in g’s) is given as

a function of time in seconds by

a=100

1−0.2t.

What is the missile’s velocity in miles per hour 1 s after

liftoff?

Problem 13.20 The airplane releases its drag para-

chute at time t=0. Its velocity is given as a function

of time by

v=80

1+0.32tm/s.

What is the airplane’s acceleration at t=3s?

Problem 13.21 How far does the airplane in Problem

Solution:

Problem 13.22 The velocity of a bobsled is v=

10tft/s. When t=2 s, the position is s=25 ft. What

is its position at t=10 s?

Solution: The equation for straight line displacement under con-

stant acceleration is

18

Problem 13.23 In September, 2003, Tony Schumacher

Solution:

Problem 13.24 The velocity of an object is v=200 −

Solution: The acceleration is

3

3t36

3

Problem 13.25 An inertial navigation system mea-

Solution:

Problem 13.26 In Example 13.3, suppose that the chee-

tah’s acceleration is constant and it reaches its top speed

of 75 mi/h in 5 s. What distance can it cover in 10 s?

Problem 13.27 The graph shows the airplane’s accel-

eration during its takeoff. What is the airplane’s velocity

when it rotates (lifts off) at t=30 s?

a

9 ft/s2

3 ft/s2

5 s0

30 s

Solution: Velocity =Area under the curve

v=1

Problem 13.28 Determine the distance traveled during

its takeoff by the airplane in Problem 13.27.

Solution: for 0 ≤t≤5s

a=6 ft/s2

20

Problem 13.29 The car is traveling at 30 mi/h when

the trafﬁc light 295 ft ahead turns yellow. The driver

takes one second to react before he applies the brakes.

(a) After he applies the brakes, what constant rate of

deceleration will cause the car to come to a stop

just as it reaches the light?

(b) How long does it take the car to travel the 295 ft?

295 ft

30 mi/h

Problem 13.30 The car is traveling at 30 mi/h when

the trafﬁc light 295 ft ahead turns yellow. The driver

takes 1 s to react before he applies the accelerator. If

the car has a constant acceleration of 5 ft/s2and the

light remains yellow for 5 s, will the car reach the light

before it turns red? How fast is the car moving when it

reaches the light?

Problem 13.31 A high-speed rail transportation sys-

tem has a top speed of 100 m/s. For the comfort of the

passengers, the magnitude of the acceleration and decel-

eration is limited to 2 m/s2. Determine the time required

for a trip of 100 km.

A plot of velocity versus time can be made and the area under the

Problem 13.32 The nearest star, Proxima Centauri, is

4.22 light years from the Earth. Ignoring relative motion

between the solar system and Proxima Centauri, sup-

pose that a spacecraft accelerates from the vicinity of

the Earth at 0.01 g (0.01 times the acceleration due to

gravity at sea level) until it reaches one-tenth the speed

of light, coasts until it is time to decelerate, then decel-

erates at 0.01 g until it comes to rest in the vicinity of

Proxima Centauri. How long does the trip take? (Light

travels at 3 ×108m/s.)

reach 0.1 times the speed of light is

The total time of ﬂight is ttotal =t1+t2+t3=1.63751 ×109sec-

22

Problem 13.33 A race car starts from rest and accel-

erates at a=5+2tft/s2for 10 seconds. The brakes

are then applied, and the car has a constant acceler-

ation a=−30 ft/s2until it comes to rest. Determine

(a) the maximum velocity, (b) the total distance trav-

eled; (c) the total time of travel.

from which

vmax =150 ft/s .

(b) The distance traveled in the ﬁrst interval is

s(10)=10

0(5t+t2)dt =5

2t2+1

3t310

0

=583.33 ft.

The time of travel in the second interval is

v(t2−10)=0=a(t2−10)+v(10), t2≥10 s,

from which

(t2−10)=−150

−30 =5,and

Problem 13.34 When t=0, the position of a point is

s=6 m and its velocity is v=2 m/s. From t=0tot=

6 s, the acceleration of the point is a=2+2t2m/s2.

From t=6 s until it comes to rest, its acceleration is

a=−4 m/s2.

(a) What is the total time of travel?

(b) What total distance does the point move?

Problem 13.35 Zoologists studying the ecology of the

run 65 km/h. If the animals run along the same straight

Solution: The top speeds are Vc=100 km/h =27.78 m/s for the

4=6.94 m/s2for the cheetah, and as=Vs

4=4.513 m/s2for the

springbuck. Divide the intervals into the acceleration phase and the

Problem 13.36 Suppose that a person unwisely drives

75 mi/h in a 55 mi/h zone and passes a police car going

55 mi/h in the same direction. If the police ofﬁcers begin

constant acceleration at the instant they are passed and

increase their speed to 80 mi/h in 4 s, how long does it

take them to be even with the pursued car?

Solution: The conversion from mi/h to ft/s is

The distance traveled by the pursued car during this acceleration is

Problem 13.37 If θ=1 rad and dθ

dt =1 rad/s, what

is the velocity of Prelative to O?

24

Problem 13.38 In Problem 13.37, if θ=1 rad, dθ/dt

=−2 rad/s and d2θ/dt2=0, what are the velocity and

Solution: The velocity is

Problem 13.39 If θ=1 rad and dθ

dt =1 rad/s, what

is the velocity of Prelative to O?

O

P

200 mm 400 mm

␪

Solution: The acute angle formed by the 400 mm arm with the

horizontal is given by the sine law:

From the expression for the angle

Problem 13.40 In Active Example 13.4, determine the

time required for the plane’s velocity to decrease from

50 m/s to 10 m/s.

Solution: From Active Example 13.4 we know that the accelera-

Problem 13.41 An engineer designing a system to con-

trol a router for a machining process models the sys-

tem so that the router’s acceleration (in in/s2) during an

interval of time is given by a=−0.4v, where vis the

velocity of the router in in/s. When t=0, the position is

s=0 and the velocity is v=2 in/s. What is the position

at t=3s?

Solution: We will ﬁrst ﬁnd the velocity at t =3.

26

Problem 13.42 The boat is moving at 10 m/s when

its engine is shut down. Due to hydrodynamic drag, its

subsequent acceleration is a=−0.05v2m/s2, where v

is the velocity of the boat in m/s. What is the boat’s

velocity 4 s after the engine is shut down?

Solution:

Problem 13.43 In Problem 13.42, what distance does

Solution: From Problem 13.42 we know

Problem 13.44 A steel ball is released from rest in a

container of oil. Its downward acceleration is a=2.4−

0.6vin/s2, where vis the ball’s velocity in in/s. What is

the ball’s downward velocity 2 s after it is released?

Solution:

Problem 13.45 In Problem 13.44, what distance does

Solution: From 13.44 we know

Problem 13.46 The greatest ocean depth yet discov-

ered is the Marianas Trench in the western Paciﬁc Ocean.

A steel ball released at the surface requires 64 minutes

to reach the bottom. The ball’s downward acceleration

is a=0.9g−cv, where g=9.81 m/s2and the constant

c=3.02 s−1. What is the depth of the Marianas Trench

in kilometers?

Solution:

Integrating,

Problem 13.47 The acceleration of a regional airliner

Solution:

Problem 13.48 In Problem 13.47, what distance does

the airliner require to take off?

Solution:

28