Question

A bus starting from rest moves with a uniform acceleration of $0.1{\mathrm{ms}}^{-2}$ for $2minutes$. Find (a) the speed acquired, and (b) the distance travelled.

Answer 1

Step 1: Given

As the bus starts from rest, so, the initial velocity $\left(u\right)=0m/s$

Acceleration $\left(a\right)=0.1m{s}^{-2}$

Time $=2minutes=120s$

Step 2: Using the first motion equation for determining the value of speed acquired

As per the first motion equation, we have, $v=u+at$

Here $v$ is the final velocity and $u$ is the initial velocity and $a$ is acceleration and $t$is the time taken.

So, the value of the speed acquired will be $\Rightarrow v=0+0.1\times 120=12m/s$

Step 3: Applying the third motion equation for determining the value of distance

As per the third motion equation, we have,

$$\begin{gathered} v^2-u^2=2as \\ \Rightarrow s=\frac{v^2-u^2}{2a} \end{gathered}$$

So, the value of distance will be $s=\frac{v^2-u^2}{2a}=\frac{{12}^2-0^2}{2\times 0.1}=\frac{144}{0.2}=720m$

Hence, the speed acquired is $12m{s}^{-1}$ and the total distance traveled is $720m$.