Question

A car travelling at a speed of $30km/h$ is brought to a halt in $8m$ by applying brakes. If the same car is travelling at $60km/h$ it can be brought to a halt with the same braking power in

• A. $8m$
• B. $16m$
• C. $24m$
• D. $32m$

Answer 1

The correct option is D

$32m$

Step 1: Given data

Speed of car, $u_1=30km/h$

$u_2=60Km/h$

Distance of halt, $s=8m$

Step 2: Find retardation when car traveling at a speed of $30km/h$

Car is brought to a halt i.e. final velocity $=0$

Using equation of motion,

$\begin{array}{rcl}v& =& u+at\end{array}$

$\Rightarrow$ $\begin{array}{rcl}u& =& –at\end{array}$

$\Rightarrow$ $\begin{array}{rcl}t& =& -\frac{u}{a}\end{array}$

By using the formula, we get

$S_1=ut+\frac{1}{2}a{t}^2$

$\Rightarrow$$S_1=\left(-at\right)t+\frac{1}{2}a{t}^2$

$\Rightarrow$$S_1=-\frac{1}{2}a{t}^2$

$\Rightarrow$$S_1=-\frac{1}{2}\frac{{\left(u_1\right)}^2}{a}$ $\left[\because u=u_1=30km/h\right]$

$\Rightarrow$$S_1=-\frac{1}{2}\frac{{\left(30\right)}^2}{a}$

Step 3: Find distance covered by car after break at speed of $60km/h$

Again using,

$S_2=ut+\frac{1}{2}a{t}^2$

$\Rightarrow$$S_2=\left(-at\right)t+\frac{1}{2}a{t}^2$

$\Rightarrow$$S_2=-\frac{1}{2}a{t}^2$

$\Rightarrow$$S_2=-\frac{1}{2}\frac{{\left(u_2\right)}^2}{a}$ $\left[\because u=u_2=60km/h\right]$

$\Rightarrow$$S_2=-\frac{1}{2}\frac{{\left(60\right)}^2}{a}$

Divide $S_1$ by $S_2$, we get

$\frac{S_1}{S_2}=\frac{-\frac{1}{2}\frac{{\left(30\right)}^2}{a}}{-\frac{1}{2}\frac{{\left(60\right)}^2}{a}}$

$\Rightarrow$$\frac{S_1}{S_2}={\left(\frac{30}{60}\right)}^2$

$\Rightarrow$$\frac{8}{S_2}={\left(\frac{1}{2}\right)}^2$

$\Rightarrow$ $S_2=8\times 4$

$\Rightarrow$ ${\mathit{S}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{32}\mathit{m}$

Hence, option D is correct.