Question

A sports car accelerates from rest to 95 km/h in 4.3 s. What is its average acceleration in $\mathrm{m}/{\mathrm{s}}^2$?

Short answer

The acceleration of the car is $6.14 \frac{\mathrm{m}}{{\mathrm{s}}^2}$.

Step-by-step solution

Step 1. Conversion of the unit and the rule to find the acceleration

The acceleration of the body is defined as the derivative of the velocity of a body with respect to time. The unit of acceleration is $\mathrm{m}/{\mathrm{s}}^2$.

Given data.

The initial velocity of the car was $v_1 = 0$as it started from rest.

The final velocity of the car is $v_2 = 95 \frac{\mathrm{km}}{\mathrm{h}}$.

Now,

$\begin{array}{c}{v}_2 = 95 \frac{\mathrm{km}}{\mathrm{h}}\\ = 95 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{1 \mathrm{km}} \times \frac{1 \mathrm{h}}{3600 \mathrm{s}}\\ = 26.4 \frac{\mathrm{m}}{\mathrm{s}}\end{array}$

The time taken by the car for this motion is $\Delta t = 4.3 \mathrm{s}$.

Rule.

You know that acceleration is

role="math" localid="1642764168067" $a = \frac{v_2 - v_1}{\Delta t}\dots \left(i\right)$.

Step 2. Calculation of the acceleration 

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

$\begin{array}{c}a = \frac{26.4 \frac{\mathrm{m}}{\mathrm{s}} -0}{4.3 \mathrm{s}}\\ = 6.14 \frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Therefore, the acceleration of the car is $6.14 \frac{\mathrm{m}}{{\mathrm{s}}^2}$.