At highway speed, a particular automobile is capable of an acceleration of about 1.8 $\mathrm{m}/{\mathrm{s}}^2$. At this rate, how long does it take to accelerate from 65 km/h to 120 km/h?

The rate with which the body changes its velocity in a given amount of time is known as acceleration. The acceleration of a body can mathematically be defined as the double derivative of the velocity with respect to time.

Given data.

The initial velocity of the automobile is $v_1 = 65 \frac{\mathrm{km}}{\mathrm{h}}$.

The final velocity of the automobile is $v_2 = 120 \frac{\mathrm{km}}{\mathrm{h}}$.

The acceleration of the automobile is $a = 1.8 \frac{\mathrm{m}}{{\mathrm{s}}^2}$.

Assumption.

Assume that it takes t time to accelerate from $65\frac{\mathrm{km}}{\mathrm{h}}$to $120\frac{\mathrm{km}}{\mathrm{h}}$.

You know that the acceleration is

role="math" localid="1642768787969" $a = \frac{v_2 - v_1}{\Delta t}\dots \left(i\right)$.

Now,

$\begin{array}{c}{v}_1 = 5 \frac{\mathrm{km}}{\mathrm{h}}\\ = 65 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{1 \mathrm{km}} \times \frac{1 \mathrm{h}}{3600 \mathrm{s}}\\ = 18.06 \frac{\mathrm{m}}{\mathrm{s}}\end{array}$,

And,

$\begin{array}{c}{v}_2 = 120 \frac{\mathrm{km}}{\mathrm{h}}\\ = 120\frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{1 \mathrm{km}} \times \frac{1 \mathrm{h}}{3600 \mathrm{s}}\\ = 33.33 \frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Now, substituting all the values of $v_1,v_2$and $a$in equation (i),

$\begin{array}{c}1.8 \frac{\mathrm{m}}{{\mathrm{s}}^2}=\frac{33.33 \frac{\mathrm{m}}{\mathrm{s}} - 18.06 \frac{\mathrm{m}}{\mathrm{s}}}{\Delta t}\\ \Delta t= \frac{33.33 \frac{\mathrm{m}}{\mathrm{s}} - 18.06 \frac{\mathrm{m}}{\mathrm{s}}}{1.8 \frac{\mathrm{m}}{{\mathrm{s}}^2} }\\ \Delta t= 8.48 \mathrm{s}\end{array}$

Therefore, it takes 8.48 s to accelerate from $65\frac{\mathrm{km}}{\mathrm{h}}$to $120\frac{\mathrm{km}}{\mathrm{h}}$.