Chapter 2, Problem 19P

To determine

The magnitude of its acceleration (assumed constant) in m/s², and in g’s ( $g=9.80{\text{ m/s}}^{\text{2}}$ ).

Answer to Problem 19P

Solution:

  $\text{Acceleration }=-6m/s^2\text{ and}-0.61g’s$

Explanation of Solution

Given:

Distance travelled by sports car at constant velocity is $120\text{m}$ in 5.0 s.

The car comes to rest in 4.0 s.

Formula Used:$\text{Speed =}\frac{\text{Distance}}{\text{Time}}$

  $\text{Acceleration}\text{=}\frac{\text{Change}\text{in}\text{velocity}}{\text{Time}\text{taken}}$

Calculation:

First calculate the initial speed.

  $u\text{=}\frac{\text{120}\text{m}}{\text{5}\text{.0}\text{s}}=24\text{m/s}$

Final speed, $v=0$

  $\begin{array}{l}t=4.0\text{ s}\\ a=\frac{v-u}{t}\\ a=\frac{\text{0-24 m/s}}{\text{4}\text{.0 s}}\\ a=-6{\text{ m/s}}^{\text{2}}\end{array}$

  $g=9.8{\text{ m/s}}^{\text{2}}$

Divide by $g=9.8{\text{ m/s}}^{\text{2}}$ to find the acceleration in g’s

  $a=\frac{-6}{9.8}=-0.61g’s$