Chapter 3, Problem 60A

(a)

To determine

To Plot:Velocity-time graph using the given table.

To Find: The time interval during which the object speeds up and slows down.

Explanation of Solution

Given:

Time (s)	Velocity (m/s)
0	4
1	8
2	12
3	14
4	16
5	16
6	14
7	12
8	8
9	4
10	0
11	-4
12	-8

Plot:

  

The object speeds up in from 0 to 4 s, then maintains constant speed between 4 s and 5 s. Thereafter, the object slows from 5 s to 10 s. The speed increases again.

(b)

To determine

To Find: The time at which the object reverses the direction.

Explanation of Solution

Introduction:

  

The object speeds up in from 0 to 4 s, then maintains constant speed between 4 s and 5 s. Thereafter, the object slows from 5 s to 10 s. The speed increases again.

At 10 s, it can be implied from the plot that the object changes the direction because the speed increase but in negative direction.

(c)

To determine

To Compare: The average acceleration between 0.0 to 2.0 s and 7.0 s and 12.0 s.

Explanation of Solution

Given:

Time (s)	Velocity (m/s)
0	4
1	8
2	12
3	14
4	16
5	16
6	14
7	12
8	8
9	4
10	0
11	-4
12	-8

Formula Used:

Average acceleration can be obtained by:

  $a_{avg}=\frac{v_f-v_i}{t}$

Here, $v_f$ is the final velocity, $v_i$ is the initial velocity and t is the time.

Calculations:

The average acceleration between 0.0 to 2.0 s:

  $\begin{array}{l}{a}_{avg}=\frac{v_f-v_i}{t}\\ a_{avg}=\frac{12-4}{2}=4{\text{m/s}}^{\text{2}}\end{array}$

The average acceleration between 7.0 s and 12.0 s:

  $\begin{array}{l}{a}_{avg}=\frac{-8-\left(-12\right)}{5}\\ a_{avg}=\frac{12-4}{2}=0.8{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

The average acceleration between 0.0 to 2.0 s is greater than that between 7.0 s and 12.0 s.