Question

A car moving with a velocity of $10m{s}^{-1}$ can be stopped by the application of constant force $F$ in a distance of $20m$. If the velocity of the car is $30m{s}^{-1}$, it can be stopped by this force in

• A. $\frac{20}{3}m$
• B. $20m$
• C. $60m$
• D. $180m$

Answer 1

The correct option is D

$180m$

Step 1. Given data

For case $1-$

The initial velocity of the car, $u_1=10m{s}^{-1}$

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

Distance, $s=20m$

For case $2-$

The initial velocity of the car, $u_2=30m{s}^{-1}$

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

Step 2. Finding the acceleration in the first case

By using the equation of motion

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

$0={\left(10\right)}^2+2\times a\times 20$

$a=-\frac{10\times 10}{2\times 20}$

$a=-2.5m{s}^{-2}$

Step 3. Finding the stopping distance

Again, using the equation of motion

$v_2^2=u_2^2+2as$

$0={\left(30\right)}^2+2\times \left(-2.5\right)\times s$

$s=\frac{30\times 30}{2\times 2.5}$

$s=180m$

Therefore, the stopping distance is $180m$

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