Chapter 4, Problem 71A

To determine

The acceleration and the final velocity of the car if it has a constant acceleration.

Answer to Problem 71A

  $a=31.88{\text{m/s}}^{\text{2}}$ and $v_f=160.133\text{m/s}$

Explanation of Solution

Given:

A dragster completed a $402.3\text{ m}$ run in $5.023\text{ s}$ .

Formula used:

Final velocity with constant acceleration can be obtained by:

  $v_{final}=v_{initial}+a\Delta t$

Where, $\Delta t$ is the time taken, a is acceleration and v represents velocity.

The distance traveled by an object with a constant acceleration can be obtained by:

  $d_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a\Delta t^2$

Here, d represents displacement, v represents velocity, t represents time and a is the acceleration.

Calculation:

Since the dragster starts from the rest, hence its initial velocity will be 0, that is

  $v_i=0\text{m/s}$

Now, using the above-mentioned formula for the distance covered, the acceleration of the motion will be:

  $\begin{array}{c}{d}_f=d_i+v_i\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ 402.3=0+\left(0\right)\left(5.023\right)+\frac{1}{2}a{\left(5.023\right)}^2\\ a=31.86{\text{m/s}}^{\text{2}}\end{array}$

Then, the final velocity of the car will be:

  $\begin{array}{c}{v}_f=v_i+a\Delta t\\ =0+\left(31.88\right)\left(5.023\right)\\ =160.133\text{m/s}\end{array}$

Conclusion:

Thus, the acceleration and the final velocity of the car will be $31.88{\text{m/s}}^2$ and $160.133\text{m/s}$ , respectively.