Chapter 3.2, Problem 34SSC

To determine

Acceleration provided by the car brakes.

Answer to Problem 34SSC

  $-1.26{\text{ m/s}}^2$

Explanation of Solution

Given:

Initial speed of the car, $u=23\text{ m/s}$

Distance covered by car before coming to stop, $s=210\text{ m}$

Formula used:

Third equation of motion:

  $v^2=u^2+2as$

Here, v is the final velocity, u is initial velocity, a is acceleration and s is displacement.

Calculation:

As the car finally stops, therefore final velocity of car $v=0$

Now, substitute the values of v , u and sand solve:

  $\begin{array}{l}{0}^2={23}^2+2\times a\times 210\\ 420a=-{23}^2\\ a=\frac{-529}{420}\\ =-1.26{\text{ m/s}}^2\end{array}$

Here, negative sign indicates that speed of the car decreases with time.

Conclusion:

Therefore, acceleration of the car will be $-1.26{\text{ m/s}}^2$ .