Mechanical Engineering Chapter 12 Homework Draw The At Graph And Determine The

4 1

12–41.

SOLUTION

Thus

Thus,

10 h=160 –4h

t1=0.4 t2

+cv2=v1+act1

The elevator starts from rest at the first floor of the

building. It can accelerate at and then decelerate at

Determine the shortest time it takes to reach a floor

40 ft above the ground. The elevator starts from rest and

then stops. Draw the a–t,v–t, and s–tgraphs for the motion.

2ft>s2.

5ft>s2

40 ft

4 2

12–42.

The velocity of a car is plotted as shown. Determine the

total distance the car moves until it stops

Construct the a–t graph. 1t=80 s2.

t(s)

10

40 80

v

(m/s)

SOLUTION

Distance Traveled:The total distance traveled can be obtained by computing the

area under the graph.

a–tGraph:The acceleration in terms of time tcan be obtained by applying .

For time interval ,

0s …t640 s

a=dv

dt

v–t

a

=–

0.250 m

>

s

4 3

SOLUTION

v–t Graph. The v–t function can be determined by integrating

dv=a dt

.

For

0…t610 s,

a=0

. Using the initial condition

v=300 ft>s

at

t=0,

>

t

L

300 ft

>

s

dv=L

t

10 s

(t–30) dt

At

t=20 s,

L

150 ft

>

s

dv=L

t

20 s

–10 dt

t=t′,

Using these results, the

v9t

graph shown in Fig. a can be plotted s-t Graph.

The

s9t

function can be determined by integrating

ds =v dt

. For

0…t610 s,

the

12–43.

The motion of a jet plane just after landing on a runway

is described by the a–t graph. Determine the time t

′

when

the jet plane stops. Construct the v–t and s–t graphs for the

motion. Here s = 0, and v = 300 ft

>

s when t = 0.

t (s)

10

a (m/s2)

10

20 t¿

20

v

4 4

12–43. Continued

For 10 s

6

t

6

20 s, the initial condition is s = 3000 ft at t = 10 s.

Ls

3000 ft

At t=20 s,

For

20 s 6t…35 s,

the initial condition is s

=

5167 ft at t

=

20 s.

Ls

5167 ft

ds =

Lt

20 s

(–10t+350) dt

At t=35 s,

using these results, the s-t graph shown in Fig. b can be plotted.

4 5

*12–44.

The v–tgraph for a particle moving through an electric field

from one plate to another has the shape shown in the figure.

The acceleration and deceleration that occur are constant

and both have a magnitude of If the plates are

spaced 200 mm apart, determine the maximum velocity

and the time for the particle to travel from one plate to

the other. Also draw the s–tgraph. When the

particle is at s=100 mm.

t=t¿>2

t¿

vmax

4m>s2.

SOLUTION

ac=4 m/s2

t¿/2t¿t

v

s

max

v

max

s

4 6

12–45.

SOLUTION

For ,

When ,

Ls

0.5

ds =Lt

0.11–100t+202dt

t=0.1 s

v=100 t

06t60.1 s

t¿/2t¿t

v

s

max

v

max

s

The v–t graph for a particle moving through an electric field

from one plate to another has the shape shown in the figure,

where t¿ = 0.2 s and vmax = 10 m>s. Draw the s–t and a–t graphs

for the particle. When t = t¿>2 the particle is at s = 0.5 m.

Ans:

4 7

SOLUTION

0…s6100

12–46.

The a–s graph for a rocket moving along a straight track has

been experimentally determined. If the rocket starts at s = 0

when v = 0, determine its speed when it is at

s = 75 ft, and 125 ft, respectively. Use Simpson’s rule with

n = 100 to evaluate v at s = 125 ft.

Ans:

s (ft)

a (ft/s2)

100

5

a  5  6(s  10)5/3

4 8

12–47.

A two-stage rocket is fired vertically from rest at s = 0 with

the acceleration as shown. After 30 s the first stage, A, burns

out and the second stage, B, ignites. Plot the v–t and s–t

graphs which describe the motion of the second stage for

0… t … 60 s.

SOLUTION

v9t

Graph. The

v–t

function can be determined by integrating

dv=a dt

.

At t=30 s,

At t=60 s,

At t=30 s,

24

12

A

B

a (m/s2)

4 9

For

30 s 6t…60 s

, the initial condition is

s=1800 m

at t = 30 s.

12–47. Continued

Ans:

5 0

SOLUTION

For

0…t64 s

*12–48.

The race car starts from rest and travels along a straight

road until it reaches a speed of 26 m

>

s in 8 s as shown on the

v–t graph. The flat part of the graph is caused by shifting

gears. Draw the a–t graph and determine the maximum

acceleration of the car. 26

14

584

t (s)

v (m/s)

v  3.5t

v  4t  6

6

Ans:

5 1

12–49.

SOLUTION

v9t

Function. The

v–t

function can be determined by integrating

dv=a dt

. For

0…t615 s

,

a=6 m>s2.

Using the initial condition

v=10 m>s

at

t=0

,

v={6t+10} m>s

The jet car is originally traveling at a velocity of 10 m

>

s

when it is subjected to the acceleration shown. Determine

the car’s maximum velocity and the time t

′

when it stops.

When t = 0, s = 0.

6

15

4

t (s)

a (

m

/s2)

t¿

Ans:

5 2

SOLUTION

For

s6300 ft

12–50.

The car starts from rest at s = 0 and is subjected to an

acceleration shown by the a–s graph. Draw the v–s graph

and determine the time needed to travel 200 ft.

s (ft)

a (ft/s2)

a  0.04s  24

300

6

12

450

5 3

v

(m

/

s)

10

6

SOLUTION

s–t Graph. The s–t function can be determined by integrating

ds =v dt

.

t=60 s

For

60 s 6t6120 s,

v=6 m>s

. Using the initial condition

s=180 m

at

t=60 s

,

Ls

180 m

ds =

Lt

60 s

6 dt

t–120

condition

s=540 m

at

t=120 s

,

Ls

Lt

t=180 s

12–51.

The v–t graph for a train has been experimentally

determined. From the data, construct the s–t and a–t graphs

for the motion for 0 … t … 180 s. When t = 0, s = 0.

5 4

12–51. Continued

a–t Graph. The a–t function can be determined using a=

dv

dt

.

0…t660 s

Using these results, a–t graph shown in Fig. b can be plotted.

Ans:

For

0…t660 s

,

5 5

*12–52.

A motorcycle starts from rest at s = 0 and travels along a

straight road with the speed shown by the v–t graph.

Determine the total distance the motorcycle travels until it

stops when t = 15 s. Also plot the a–t and s–t graphs.

5

10 154

t (s)

v (m/s)

v  1.25tv  5

v  t  15

SOLUTION

For

t64 s

When

t=4 s,

s=10 m

Ans:

SOLUTION

At t=8 s

At t=12 s

12–53.

A motorcycle starts from rest at s = 0 and travels along a

straight road with the speed shown by the v–t graph.

Determine the motorcycle’s acceleration and position when

t = 8 s and t = 12 s.

5

10 154

t (s)

v (m/s)

v  1.25tv  5

v  t  15

At t=8 s

At t=12 s

5 7

12–54.

The v–t graph for the motion of a car as it moves along a

straight road is shown. Draw the s–t and a–t graphs. Also

determine the average speed and the distance traveled for

the 15-s time interval. When t = 0, s = 0.

SOLUTION

s–t Graph. The s–t function can be determined by integrating

ds =v dt

.

For

0…t65 s

,

v=0.6t2

. Using the initial condition s

=

0 at t

=

0,

s=25 m

at

t=5 s

,

Ls

25 m

ds =

Lt

5 s

1

2

(45 –3t)dt

515

15

v  0.6t2

t (s)

v (m/s)

5 8

12–54. Continued

a–t Graph. The a–t function can be determined using a=

0…t65 s

dv

.

Ans:

For

0…t65 s

,

5 9

12–55.

SOLUTION

v0=110 ft>s

An airplane lands on the straight runway,originally traveling

at 110 ft s when If it is subjected to the decelerations

shown, determine the time needed to stop the plane and

construct the s–t graph for the motion.

t¿

s=0.>

t(s)

5

a(ft/s2)

–3

15 20 t’

–8

Ans:

t′=33.3

s

6 0

SOLUTION

v9s

Function. The

v–s

function can be determined by integrating

v dv=a ds

.

At

s=50 ft,

At

s=100 ft,

L

v

37.42 ft

>

s

v dv =L

s

100 ft

a

–

3

25 s +18

b

ds

6

8

100 150

s (ft)

a (ft/s2)

*12–56.

Starting from rest at s = 0, a boat travels in a straight line

with the acceleration shown by the a–s graph. Determine

the boat’s speed when s = 50 ft, 100 ft, and 150 ft.

Ans: