Question

A car moving at speed $50km/h$ one applying brake stops at a distance of $6m$. If same car move with the velocity $100km/h$, what distance it cover after applying the brakes.

• A. $12m$
• B. $18m$
• C. $24m$
• D. $6m$

Answer 1

The correct option is C

$24m$

Step 1. Given data

For case $1$-

The initial speed of the car, $u_1$$=50km{h}^{-1}$

$u_1=13.88m{s}^{-1}$

The final speed of the car, $v_1$$=0$

Distance, $s$$=6m$

For case $2$-

The initial velocity of the car, $u_2=100km{h}^{-1}$

$u_2=27.77m{s}^{-1}$

The final velocity of the car, $v_2=0$

Step 2. Finding the acceleration $a$ in case $2$

By using the equation of motion

$v_1^2=u_1^2+2as$

$0={\left(13.88\right)}^2+2\times a\times 6$

$a=-\frac{13.88\times 13.88}{2\times 6}$

$a=-16.05m{s}^{-2}$

Step 3. Finding the minimum stopping distance

Again, using the equation of motion

${v^2}_2=u^2+2as$

$0={\left(27.77\right)}^2+2\times \left(-16.05\right)\times s$

$s=\frac{27.77\times 27.77}{2\times 16.05}$

$=24.02m$

By rounding off,$s=24m$

Therefore, the minimum stopping distance is $24m$

Hence the correct option is $\left(C\right)$