Question

A car slows down from 28 m/s to rest at a distance of 88 m. What is its acceleration, assumed constant?

Short answer

The acceleration of the car is $-4.45 \frac{\mathrm{m}}{{\mathrm{s}}^2}$.

Step-by-step solution

Step 1. Relation between the initial velocity, final velocity, distance, and acceleration

Distance and displacement are used to measure the same quantities and have the same units, but displacement is defined as the shortest distance between two points.

Given data.

The initial velocity is $v_o = 28 \frac{\mathrm{m}}{\mathrm{s}}$.

The final velocity is $v=0$as comes to rest.

The distance is $\left(x - x_o\right) = 88 \mathrm{m}$.

Assumption.

Let the acceleration be $a$.

Now, by the third equation of motion,

role="math" localid="1642829767845" $v^2 = v_o^2 + 2a\left(x - x_o\right)\dots \left(i\right)$.

Step 2. Calculation of the acceleration

Now, using equation (i),

$\begin{array}{c}{v}^2 = v_o^2 + 2a\left(x - x_o\right)\\ 0^2 = {\left(28 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2 + 2a \times 88 \mathrm{m}\\ \mathrm{a} = - 4.45 \frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Therefore, the acceleration of the car is $- 4.45 \frac{\mathrm{m}}{{\mathrm{s}}^2}$.