Question

A car moving with a speed of $40km{h}^{-1}$ can be stopped by applying brakes after at least $2m$. If the same car is moving with a speed of $80km{h}^{-1}$., what is the minimum stopping distance

• A. $8m$
• B. $2$$m$
• C. $4m$
• D. $6m$

Answer 1

The correct option is A

$8m$

Step 1. Given data

For case $1$$-$

Initial velocity of the car, $u_1=40km{h}^{-1}$

$u_1=11.11m{s}^{-1}$

Final velocity of the car, $v_1=0$

Distance$=2m$

For case $2-$

Initial velocity of the car, $u_2=80km{h}^{-1}$

$u_2=22.22m{s}^{-1}$

Final velocity of the car, $v_2=0$

Step 2. Finding the acceleration in the first case

By using the equation of motion

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

$0={\left(11.11\right)}^2+2\times a\times 2$

$a=-\frac{11.11\times 11.11}{2\times 2}$

$a=-30.85m{s}^{-2}$

STEP 3. Finding the minimum stopping distance

Again, using the equation of motion

$v_2^2=u_2^2-2as$

$0={\left(22.22\right)}^2+2\times \left(-30.85\right)\times s$

$s=\frac{22.22\times 22.22}{2\times 30.85}$

$s=8m$

Therefore, the minimum stopping distance is $8m$

Hence, the correct option is $\left(A\right)$