Question

A bus moves by $8m$ from its position of rest in $2s$, along a straight road. If it covers $10m$ in the next second, it is moving with ___.

Answer 1

Step 1: Given Data

Initial velocity, $\mathrm{u}=0m{s}^{-1}$.

Distance traveled in $t_1=2s$ is $d_1=8m$.

Distance traveled in the next $t_2=1s$ is $d_2=10m$.

The diagrammatic representation of the given data is given below:

Step 2: Find the acceleration of the bus just after covering $\mathbf{8}\mathbf{}\mathit{m}$

Assume the acceleration of the bus as $am{s}^{-2}$.

Using the third equation of motion $\mathit{s}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathit{t}\mathbf{}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$ we get:

$\Rightarrow d_1=u{t}_1+\frac{1}{2}a{t_1}^2$

$$\begin{gathered} \Rightarrow 8=0+\hspace{0.17em}\frac{a}{2}\times 2^2 \\ \Rightarrow 8=2a \\ \Rightarrow a=\frac{8}{2} \\ \Rightarrow a=4m{s}^{-2} \end{gathered}$$

Step 3: Find the acceleration of the bus just after covering $\mathbf{10}\mathbf{}\mathit{m}$

The total distance covered by bus, $s=\left(d_1+d_2\right)=\left(8+10\right)=18m$

Total time taken by the bus, $t=\left(t_1+t_2\right)=\left(2+1\right)=3s$

Using the third equation of motion $\mathit{s}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathit{t}\mathbf{}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$ we get:

$$\begin{gathered} \Rightarrow 18=0+\hspace{0.17em}\frac{a}{2}\times 3^2 \\ \Rightarrow 18=\frac{9a}{2} \\ \Rightarrow a=\frac{18\times 2}{9} \\ \Rightarrow a=4m{s}^{-2} \end{gathered}$$

As the magnitude and direction of the accelerations of the bus are the same at $2s$ and $3s$, it is moving with a constant acceleration.

Therefore, the bus is moving at a constant acceleration of $\mathbf{4}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{2}}$.