Question

A bus starts moving with uniform acceleration from its position of rest. It moves $48m$ in $4s$. On applying the brakes, it stops after covering $24m$. Find the deceleration of the bus.

Answer 1

Step 1: Given data

A bus starts from rest which means the initial velocity of the bus is $u=0m{s}^{-1}$

Distance covered in $4s$ with uniform acceleration is $s_1=48m$.

Assume the uniform acceleration as $a_{1}m{s}^{-2}$.

On applying brakes, it stops after covering distance $s_2=24m$.

The final velocity of the bus $v=0m{s}^{-1}$.

Assume the deceleration as $-a_{2}m{s}^{-2}$.

The negative sign represents that the direction of the deceleration applied is in the opposite direction of the motion of the bus.

Step 2: Find the velocity of the bus at the end of $\mathbf{4}\mathbf{}\mathit{s}$

Since we know the second equation of the motion is $\mathit{s}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathit{t}\mathbf{}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$

To find the acceleration of the bus in $4s$ we have to put all the required data in the above equation,

$\Rightarrow s_1=ut+\frac{1}{2}{a}_1{t}^2$

$\Rightarrow 48=0+\frac{1}{2}{a}_1{\left(4\right)}^2$

$\Rightarrow 48=\frac{1}{2}\times a_1\times 16$

$\Rightarrow a_1=\frac{48\times 2}{16}$

$\Rightarrow a_1=6m{s}^{-2}$

Since we know the first equation of motion is $\mathit{v}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathbf{}\mathbf{+}\mathbf{}\mathit{a}\mathit{t}$

$\Rightarrow v=u+a_{1}t$

$\Rightarrow v=0+\left(6\right)\left(4\right)$

$\Rightarrow v=24m{s}^{-1}$

Step 3: Find the deceleration of the bus after applying brakes

On applying brakes the distance covered by bus is $s_2=24m$.

The velocity of the bus at $4s$ becomes the initial velocity of the bus during this time interval.

Therefore, the initial velocity of the bus before applying the brakes is $u=24m{s}^{-1}$.

The final velocity of the bus is $v=0m{s}^{-1}$ as it stops at the end of $24m$.

Since we know the third equation of motion is ${\mathit{v}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{u}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathit{a}\mathit{s}$.

On applying the given data we get,

$\Rightarrow 0^2={24}^2+2\left(-a_2\right)\left(24\right)$

$\Rightarrow 0=576-48a_2$

$$\begin{gathered} \Rightarrow 48a_2=576 \\ \Rightarrow a_2=\frac{576}{48} \end{gathered}$$

$\Rightarrow a_2=12m{s}^{-2}$

Hence, the deceleration of the bus is $\mathbf{12}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{2}}$.