Chapter 4, Problem 8A

By how much does the speed of a vehicle moving in a straight line change each second when it is accelerating at $\text{2 km/h}\cdot \text{s}$? At $\text{4 km/h}\cdot \text{s}$? At $\text{10 km/h}\cdot \text{s}$?

To determine

The change in the speed of vehicle each second.

Explanation of Solution

Given:

Acceleration, $a_1=2\frac{\text{km}}{\text{h}\cdot \text{s}}$

Time interval $=t=1\text{ s}$

Change in speed $=\Delta v_1$

Acceleration, $a_2=4\frac{\text{km}}{\text{h}\cdot \text{s}}$

Change in speed $=\Delta v_2$

Acceleration, $a_3=10\frac{\text{km}}{\text{h}\cdot \text{s}}$

Change in speed $=\Delta v_3$

Formula Used:

According to kinematics equation, change in speedis:

  $\Delta v=at$

Calculation:

When acceleration is $a_1=2\frac{\text{km}}{\text{h}\cdot \text{s}}$

Change in speed is given as

  $\begin{array}{l}\Delta v_1=a_{1}t\\ \Delta v_1=\left(\text{2}\frac{\text{km}}{\text{h}\cdot \text{s}}\right)\left(1\text{ s}\right)\\ \Delta v_1=2\frac{\text{km}}{\text{h}}\end{array}$

When acceleration is $a_2=4\frac{\text{km}}{\text{h}\cdot \text{s}}$

Change in speed is given as

  $\begin{array}{l}\Delta v_2=a_{2}t\\ \Delta v_2=\left(4\frac{\text{km}}{\text{h}\cdot \text{s}}\right)\left(1\text{ s}\right)\\ \Delta v_2=4\frac{\text{km}}{\text{h}}\end{array}$

When acceleration is $a_3=10\frac{\text{km}}{\text{h}\cdot \text{s}}$ .

Change in speed is given as

  $\begin{array}{l}\Delta v_3=a_{3}t\\ \Delta v_3=\left(10\frac{\text{km}}{\text{h}\cdot \text{s}}\right)\left(1\text{ s}\right)\\ \Delta v_3=10\frac{\text{km}}{\text{h}}\end{array}$

Conclusion:

Hence, the change in speed comes out to be $\text{2}\frac{\text{km}}{\text{h}},4\frac{\text{km}}{\text{h}}\text{and}\text{10}\frac{\text{km}}{\text{h}}$ .