Question

A car slows down uniformly from a speed of 28.0 m/s to rest in 8.0 s. How far does it travel in this time?

Short answer

The car travels 112.0 m in the given time.

Step-by-step solution

Step 1. Relation between initial velocity, final velocity, acceleration, and displacement

The velocity of a body is the vector quantity and has the same unit as that of speed. It is defined as the rate of change of displacement in unit time.

Given data.

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

The final speed of the car is $v = 0$as it comes to rest.

The time taken to come to rest is $\Delta t = 8.0 \mathrm{s}$.

Assumption.

Let the acceleration be a.

Also, assume that it travels $\Delta x$distance in the given time.

Now, you know that

role="math" localid="1642836157076" $v = v_o + a \Delta t\dots \left(i\right)$.

The second equation of motion is

role="math" localid="1642836200507" $v^2 = v_o^2 + 2a \Delta x\dots \left(ii\right)$.

Step 2. Calculation of the acceleration

Now, substituting the values in equation (i),

$\begin{array}{c}v = v_o + a \Delta t\\ 0 = 28 \frac{\mathrm{m}}{\mathrm{s}} +\left(a \times 8.0 \mathrm{s}\right)\\ a = - 3.50 \frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Step 3. Calculation of the traveled distance

Now, substituting all the values in equation (ii),

$\begin{array}{c}{v}^2 = v_o^2 + 2a \Delta x\\ 0^2 = {\left(28.0 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2 + \left[2 \times \left(- 3.50 \frac{\mathrm{m}}{{\mathrm{s}}^2}\right) \times \Delta x\right]\\ 7.00 \frac{\mathrm{m}}{{\mathrm{s}}^2} \times \Delta x = {\left(28.0 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2\\ \Delta x = 112.0 \mathrm{m}\end{array}$

Therefore, the car travels 112.0 m in the given time.