Chapter 3, Problem 92A

To determine

The distance travelled by the car before coming to stop.

Answer to Problem 92A

  $14.58\text{m}$

Explanation of Solution

Given:

The initial velocity of the car is $v_{\text{i}}=110\text{km/h}$ .

After hitting the barrier, the car’s velocity decreases at the rate, $\overline{a}=-32{\text{m/s}}^2$ .

Consider the initial distance covered by the car is $d_{\text{i}}=0\text{km}$ .

Consider the final distance covered by the car before coming to stop is $d_{\text{f}}$ .

The final velocity of the car is $v_{\text{f}}=0\text{km/h}$ .

Formula used:

Equation of motion for uniform acceleration: Consider an object is moving with uniform acceleration, then the equation of motion can be given as follows:

  $v_{\text{f}}^2-v_{\text{i}}^2=2\overline{a}\left(d_{\text{f}}-d_{\text{i}}\right)$

Here, $v_{\text{i}}$ and $v_{\text{f}}$ are the initial and final velocity of the object, $d_{\text{i}}$ and $d_{\text{f}}$ are the initial and final distance covered by the object, and $\overline{a}$ is the acceleration of the object.

Calculation:

Find the distance covered by the car before coming to stop as follows:

Apply the equation of motion,

  $\begin{array}{c}{v}_{\text{f}}^2-v_{\text{i}}^2=2\overline{a}\left(d_{\text{f}}-d_{\text{i}}\right)\\ 0^2-{\left(110\text{km/h}\times \frac{0.2777\text{m/s}}{1\text{km/h}}\right)}^2=2\times \left(-32\right)\left(d_{\text{f}}-0\right)\\ 0^2-{\left(30.547\text{m/s}\right)}^2=2\times \left(-32\right)\left(d_{\text{f}}-0\right)\\ -933.1192=-64d_{\text{f}}\\ d_{\text{f}}=\left(\frac{933.1192}{64}\right)\\ d_{\text{f}}=14.58\text{m}\end{array}$

Conclusion:

Thus, the distance traveled by car before coming to stop is $14.58\text{m}$ .