Chapter 3, Problem 95A

(a)

To determine

The velocity time graph of the motion.

Answer to Problem 95A

The velocity time graph of the motion is given in the figure (1).

Explanation of Solution

The velocity time graph of the motion is given in the figure (1).

  

Figure (1)

(b)

To determine

The displacement of the car in first 2 s.

Answer to Problem 95A

The displacement of the car in first 2 s is $8\text{m}$ .

Explanation of Solution

Given Info:

The time to for the displacement is, $t=2\text{s}$

The velocity corresponding to the time 2 s is, $v=8\text{m}/\text{s}$

Formula Used:

The expression to calculate the displacement is,

  $d=\frac{1}{2}vt$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}d=\frac{1}{2}\left(8\text{m}/\text{s}\right)\left(2\text{s}\right)\\ =8\text{m}\end{array}$

Thus, the displacement of the car in first 2 s is $8\text{m}$ .

(c)

To determine

The displacement of the car in first 4 s.

Answer to Problem 95A

The displacement of the car in first 4 s is $32\text{m}$ .

Explanation of Solution

Given Info:

The time to for the displacement is, $t=4\text{s}$

The velocity corresponding to the time 4 s is, $v=16\text{m}/\text{s}$

Formula Used:

The expression to calculate the displacement is,

  $d=\frac{1}{2}vt$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}d=\frac{1}{2}\left(16\text{m}/\text{s}\right)\left(4\text{s}\right)\\ =32\text{m}\end{array}$

Thus, the displacement of the car in first 4 s is $32\text{m}$ .

(d)

To determine

The displacement of the car in 8 s.

Answer to Problem 95A

The displacement of the car in first 8 s is $110\text{m}$ .

Explanation of Solution

Given Info:

The time up to which the velocity of the car increases is, $t_1=5\text{s}$

The time in which the velocity of the car remain constant is, $t_2=3$

The velocity of the car is, $v=20\text{m}/\text{s}$

Formula Used:

The expression to calculate the displacement is,

  $\begin{array}{c}d=\frac{1}{2}v{t}_1+v{t}_2\\ d=\frac{\left(t_1+2t_2\right)v}{2}\end{array}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}d=\frac{\left[5\text{s}+2\left(3\text{s}\right)\right]\left(20\text{m}/\text{s}\right)}{2}\\ =110\text{m}\end{array}$

Thus, the displacement of the car in first 8 s is $110\text{m}$ .

(e)

To determine

The slope of the line between the time $t_1=0\text{s}$ to $t_2=4\text{s}$ .

Answer to Problem 95A

The slope of the line between the time $t_1=0\text{s}$ to $t_2=4\text{s}$ is $4\text{m}/{\text{s}}^2$ and it represents the acceleration of the car.

Explanation of Solution

Given info:

The velocity for the time $t_1=0\text{s}$ is, $v_1=0\text{m}/\text{s}$ .

The velocity for the time $t_2=4\text{s}$ is, $v_2=16\text{m}/\text{s}$ .

Formula Used:

The expression to calculate the slope of the line is,

  $a=\frac{v_2-v_1}{t_2-t_1}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}a=\frac{16\text{m}/\text{s}-0\text{m}/\text{s}}{4\text{s}-0\text{s}}\\ =4\text{m}/{\text{s}}^2\end{array}$

The slope of the line is positive and it represents the acceleration of the car.

(f)

To determine

The slope of the line between the time $t_1=5\text{s}$ to $t_2=7\text{s}$ .

Answer to Problem 95A

The slope of the line between the time $t_1=5\text{s}$ to $t_2=7\text{s}$ is $0\text{m}/{\text{s}}^2$ and it represent that the car is moving with constant velocity.

Explanation of Solution

Given info:

The velocity for the time $t_1=5\text{s}$ is, $v_1=20\text{m}/\text{s}$ .

The velocity for the time $t_2=7\text{s}$ is, $v_2=20\text{m}/\text{s}$ .

Formula Used:

The expression to calculate the slope of the line is,

  $a=\frac{v_2-v_1}{t_2-t_1}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}a=\frac{20\text{m}/\text{s}-20\text{m}/\text{s}}{7\text{s}-5\text{s}}\\ =0\text{m}/{\text{s}}^2\end{array}$

The slope of the line is constant and it represent that the car is moving with constant velocity.