Chapter 3.2, Problem 28PP

To determine

To Calculate: The displacement of the car.

Answer to Problem 28PP

  $31.48\text{m}$

Explanation of Solution

Given:

The car has an initial velocity of $24.\text{5 m/s}$ east while a constant acceleration of $4.2{\text{m/s}}^{\text{2}}$ in the west direction.

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:

First find the time elapsed:

  $\begin{array}{c}{v}_{final}=v_{initial}+a\Delta t\\ 18.3=24.5+\left(-4.2\right)\Delta t\\ \Delta t=1.47\text{s}\end{array}$

Negative sign of acceleration is used since it is acted towards west direction, opposite to the direction of motion of the car.

Now, assume that the initial position of the car is $d_{initial}=0\text{m}$ , then

  $\begin{array}{c}{d}_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ d_{final}=0+24.5\left(1.47\right)+\frac{1}{2}\left(-4.2\right){\left(1.47\right)}^2\\ d_{final}=36.015-4.537\\ =31.48\text{m}\end{array}$

Conclusion:

Thus, displacement of the car is $31.48\text{m}$ .