91Ó°ÊÓ

Distance, s = 0.1 Km

Time, t = 8s

Post distance = 0.1 Km

Acceleration is defined as the increase in speed or rate of any motion.

The kinematic equations of motion for a particle can be expressed as follow:

$\begin{array}{l}v=u+\mathrm{Î\pm }t\\ s=S_0+ut+\frac{1}{2}\mathrm{Î\pm }{t}^2\\ v^2-u^2=2\mathrm{Î\pm }\mathrm{s}\end{array}$

Here, s is the distance travelled, $s_0$is the initial distance of the particle, u is the initial velocity, $\mathrm{Î\pm }$is the acceleration, v is the final velocity, and t is the time taken.

Calculate the acceleration of the car as follows:

From the given data, the car is at rest initially.

That is, u = 0 and $s_0=0$

And the car covers a distance s = 0.1 Km in time t = 8 s .

Rearrange the equation $S=ut+\frac{1}{2}\mathrm{Î\pm }{t}^2$for $\mathrm{Î\pm }$as follows:

$\begin{array}{l}s=ut+\frac{1}{2}\mathrm{Î\pm }{t}^2\\ \frac{1}{2}\mathrm{Î\pm }{t}^2=s-ut\\ \mathrm{Î\pm }=\frac{2(s-ut)}{t^2}\end{array}$

Substitute 0m/s for u , 0.1 km for s , and 8s for t in equation $\mathrm{Î\pm }=\frac{2\left(S-ut\right)}{t^2}$

$$\begin{gathered} \mathrm{Î\pm }=\frac{2\left[0.1Km\left({\displaystyle \frac{1000m}{1km}}\right)-\left(0m/s\right)\left(8s\right)\right]}{{\left(8s\right)}^2} \\ \mathrm{Î\pm }=3.125m/s^2 \\ \mathrm{Î\pm }=3m/s^2 \end{gathered}$$

Therefore, the acceleration of the car is $3m/s^2$.