Chapter 2, Problem 1Q

To determine

The greatest magnitude of acceleration among all the cars.

Answer to Problem 1Q

The magnitude of the acceleration of car C is greatest among all the cars.

Explanation of Solution

Given:

The time for all the cars is, $t=1\text{s}$

The initial velocity for car A is, $v_{x0}=0\text{m}/\text{s}$

The initial velocity for car B is, $v_{x0}=20\text{m}/\text{s}$

The initial velocity for car C is, $v_{x0}=10\text{m}/\text{s}$

The final velocity for car A is, $v_x=2\text{m}/\text{s}$

The final velocity for car B is, $v_x=23\text{m}/\text{s}$

The final velocity for car C is, $v_x=5\text{m}/\text{s}$

Formula used:

The expression for the acceleration of the car is given by,

  $a_x=\frac{v_x-v_{x0}}{t}$

Here, $t$ is the time interval.

Calculation:

The magnitude for the acceleration of car A is calculated as,

  $\begin{array}{c}{a}_x=\frac{v_x-v_{x0}}{t}\\ =\frac{2\text{m}/\text{s}-0\text{m}/\text{s}}{1\text{s}}\\ =2\text{m}/{\text{s}}^2\end{array}$

The magnitude for the acceleration of car B is calculated as,

  $\begin{array}{c}{a}_x=\frac{v_x-v_{x0}}{t}\\ =\frac{23\text{m}/\text{s}-20\text{m}/\text{s}}{1\text{s}}\\ =3\text{m}/{\text{s}}^2\end{array}$

The magnitude for the acceleration of the car C is calculated as,

  $\begin{array}{c}{a}_x=\frac{v_x-v_{x0}}{t}\\ =\frac{5\text{m}/\text{s}-10\text{m}/\text{s}}{1\text{s}}\\ =-5\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the magnitude of the acceleration of car C is greatest among all the cars.