Chapter 3.2, Problem 24PP

To determine

To Calculate: The distance travelled by the car in 11s.

Answer to Problem 24PP

  $\text{363}\text{m}$

Explanation of Solution

Given info:

The race car moves with an initial velocity of $44\text{ m/s}$ and slows down at a constant acceleration to a velocity of $22\text{ m/s}$ in a time period of $\text{11 s}\text{.}$

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:

The acceleration of the car will be

  $\begin{array}{c}{v}_{final}=v_{initial}+a\Delta t\\ 22=44+a\left(11\right)\\ a=\frac{22-44}{11}\\ =-2{\text{m/s}}^2\end{array}$

Negative sign denotes deceleration. Now, assuming the initial position of the car to be 0, that is,

  $d_{initial}=0\text{m}$

Then, the distance travelled by the car in 11s will be:

  $\begin{array}{c}{d}_{final}=0+\left(44\right)\left(11\right)+\frac{1}{2}\left(-2\right){\left(11\right)}^2\\ =484-121\\ =363\mathrm{m}\end{array}$

Conclusion:

Thus, the car travels $363\mathrm{m}$ in $\text{11 s}\text{.}$ .