The equation for x ( t ) for each interval of constant acceleration motion for the object whose position as a function of time.

Question

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Answer 1

Question

Chapter 2, Problem 33QAP

To determine

The equation for x(t) for each interval of constant acceleration motion for the object whose position as a function of time.

Expert Solution & Answer

Answer to Problem 33QAP

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

Explanation of Solution

Introduction:

We can determine the equation for x(t) at each time interval by plotting a graph of position versus time and calculating the slope and the intercepts manually or using a computer program like MS Excel.

t ( s )	x ( m )
0	-12
1	-6
2	0
3	6
4	12
5	15
6	15
7	15
8	15
9	18
10	24
11	33
12	45
13	60
14	65
15	70
16	75
17	80
18	85
19	90
20	95
21	100
22	90
23	80
24	70
25	70

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 1

Between t=0 s and t=4 s

x(t)=12 m−(−12 m)4 st−12 m x(t)=6t−12

Between t=4 s and t=5 s

x(t)=15 m−(12 m)1stx(t)=3t

Between t=5 s and t=8 s

x(t)=15 m

Between t=8 s and t=13 s, there is a quadratic function.

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 2

Therefore between t=8 s and t=13 s ;

x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ;

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 3

Therefore between t=13 s and t = 21 s ;

x(t)=5t−5

Between t=21 s and t=24 s

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 4

Therefore between t=21 s and t=24 s ;

x(t)=−10t+310

Between t=24 s and t=25 s ;

x(t)=70 m

Conclusion:

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

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Answer 2

Question

Chapter 2, Problem 33QAP

To determine

The equation for x(t) for each interval of constant acceleration motion for the object whose position as a function of time.

Expert Solution & Answer

Answer to Problem 33QAP

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

Explanation of Solution

Introduction:

We can determine the equation for x(t) at each time interval by plotting a graph of position versus time and calculating the slope and the intercepts manually or using a computer program like MS Excel.

t ( s )	x ( m )
0	-12
1	-6
2	0
3	6
4	12
5	15
6	15
7	15
8	15
9	18
10	24
11	33
12	45
13	60
14	65
15	70
16	75
17	80
18	85
19	90
20	95
21	100
22	90
23	80
24	70
25	70

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 1

Between t=0 s and t=4 s

x(t)=12 m−(−12 m)4 st−12 m x(t)=6t−12

Between t=4 s and t=5 s

x(t)=15 m−(12 m)1stx(t)=3t

Between t=5 s and t=8 s

x(t)=15 m

Between t=8 s and t=13 s, there is a quadratic function.

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 2

Therefore between t=8 s and t=13 s ;

x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ;

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 3

Therefore between t=13 s and t = 21 s ;

x(t)=5t−5

Between t=21 s and t=24 s

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 4

Therefore between t=21 s and t=24 s ;

x(t)=−10t+310

Between t=24 s and t=25 s ;

x(t)=70 m

Conclusion:

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

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Answer 3

Question

Chapter 2, Problem 33QAP

To determine

The equation for x(t) for each interval of constant acceleration motion for the object whose position as a function of time.

Expert Solution & Answer

Answer to Problem 33QAP

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

Explanation of Solution

Introduction:

We can determine the equation for x(t) at each time interval by plotting a graph of position versus time and calculating the slope and the intercepts manually or using a computer program like MS Excel.

t ( s )	x ( m )
0	-12
1	-6
2	0
3	6
4	12
5	15
6	15
7	15
8	15
9	18
10	24
11	33
12	45
13	60
14	65
15	70
16	75
17	80
18	85
19	90
20	95
21	100
22	90
23	80
24	70
25	70

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 1

Between t=0 s and t=4 s

x(t)=12 m−(−12 m)4 st−12 m x(t)=6t−12

Between t=4 s and t=5 s

x(t)=15 m−(12 m)1stx(t)=3t

Between t=5 s and t=8 s

x(t)=15 m

Between t=8 s and t=13 s, there is a quadratic function.

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 2

Therefore between t=8 s and t=13 s ;

x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ;

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 3

Therefore between t=13 s and t = 21 s ;

x(t)=5t−5

Between t=21 s and t=24 s

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 4

Therefore between t=21 s and t=24 s ;

x(t)=−10t+310

Between t=24 s and t=25 s ;

x(t)=70 m

Conclusion:

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

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Answer 4

Question

Chapter 2, Problem 33QAP

To determine

The equation for x(t) for each interval of constant acceleration motion for the object whose position as a function of time.

Expert Solution & Answer

Answer to Problem 33QAP

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

Explanation of Solution

Introduction:

We can determine the equation for x(t) at each time interval by plotting a graph of position versus time and calculating the slope and the intercepts manually or using a computer program like MS Excel.

t ( s )	x ( m )
0	-12
1	-6
2	0
3	6
4	12
5	15
6	15
7	15
8	15
9	18
10	24
11	33
12	45
13	60
14	65
15	70
16	75
17	80
18	85
19	90
20	95
21	100
22	90
23	80
24	70
25	70

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 1

Between t=0 s and t=4 s

x(t)=12 m−(−12 m)4 st−12 m x(t)=6t−12

Between t=4 s and t=5 s

x(t)=15 m−(12 m)1stx(t)=3t

Between t=5 s and t=8 s

x(t)=15 m

Between t=8 s and t=13 s, there is a quadratic function.

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 2

Therefore between t=8 s and t=13 s ;

x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ;

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 3

Therefore between t=13 s and t = 21 s ;

x(t)=5t−5

Between t=21 s and t=24 s

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 4

Therefore between t=21 s and t=24 s ;

x(t)=−10t+310

Between t=24 s and t=25 s ;

x(t)=70 m

Conclusion:

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 2, Problem 33QAP

To determine

The equation for x(t) for each interval of constant acceleration motion for the object whose position as a function of time.

Expert Solution & Answer

Answer to Problem 33QAP

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

Explanation of Solution

Introduction:

We can determine the equation for x(t) at each time interval by plotting a graph of position versus time and calculating the slope and the intercepts manually or using a computer program like MS Excel.

t ( s )	x ( m )
0	-12
1	-6
2	0
3	6
4	12
5	15
6	15
7	15
8	15
9	18
10	24
11	33
12	45
13	60
14	65
15	70
16	75
17	80
18	85
19	90
20	95
21	100
22	90
23	80
24	70
25	70

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 1

Between t=0 s and t=4 s

x(t)=12 m−(−12 m)4 st−12 m x(t)=6t−12

Between t=4 s and t=5 s

x(t)=15 m−(12 m)1stx(t)=3t

Between t=5 s and t=8 s

x(t)=15 m

Between t=8 s and t=13 s, there is a quadratic function.

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 2

Therefore between t=8 s and t=13 s ;

x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ;

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 3

Therefore between t=13 s and t = 21 s ;

x(t)=5t−5

Between t=21 s and t=24 s

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 4

Therefore between t=21 s and t=24 s ;

x(t)=−10t+310

Between t=24 s and t=25 s ;

x(t)=70 m

Conclusion:

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

Answer 6

Chapter 2, Problem 33QAP

To determine

The equation for x(t) for each interval of constant acceleration motion for the object whose position as a function of time.

Expert Solution & Answer

Answer to Problem 33QAP

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

Explanation of Solution

Introduction:

We can determine the equation for x(t) at each time interval by plotting a graph of position versus time and calculating the slope and the intercepts manually or using a computer program like MS Excel.

t ( s )	x ( m )
0	-12
1	-6
2	0
3	6
4	12
5	15
6	15
7	15
8	15
9	18
10	24
11	33
12	45
13	60
14	65
15	70
16	75
17	80
18	85
19	90
20	95
21	100
22	90
23	80
24	70
25	70

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 1

Between t=0 s and t=4 s

x(t)=12 m−(−12 m)4 st−12 m x(t)=6t−12

Between t=4 s and t=5 s

x(t)=15 m−(12 m)1stx(t)=3t

Between t=5 s and t=8 s

x(t)=15 m

Between t=8 s and t=13 s, there is a quadratic function.

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 2

Therefore between t=8 s and t=13 s ;

x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ;

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 3

Therefore between t=13 s and t = 21 s ;

x(t)=5t−5

Between t=21 s and t=24 s

COLLEGE PHYSICS, Chapter 2, Problem 33QAP , additional homework tip 4

Therefore between t=21 s and t=24 s ;

x(t)=−10t+310

Between t=24 s and t=25 s ;

x(t)=70 m

Conclusion:

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

Answer 7

Chapter 2, Problem 33QAP

To determine

The equation for x(t) for each interval of constant acceleration motion for the object whose position as a function of time.

Answer 8

Expert Solution & Answer

Answer to Problem 33QAP

Between t=0 s and t=4.5 s ; x(t)=6t−12

Between t=4 s and t=5 s ; x(t)=3t

Between t=5 s and t=8 s ; x(t)=15 m

Between t=8 s and t=13 s ; x(t)=1.5t2−22.5t+99

Between t=13 s and t = 21 s ; x(t)=5t−5

Between t=21 s and t=24 s ; x(t)=−10t+310

Between t=24 s and t=25 s ; x(t)=70 m

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Introduction:

We can determine the equation for x(t) at each time interval by plotting a graph of position versus time and calculating the slope and the intercepts manually or using a computer program like MS Excel.