Question

An automobile travelling at a speed of $60km/h$, can brake to stop within a distance of $20m$. if the car is going twice as fast, i.e. $120km/h$, the stopping distance will be

• A. $20m$
• B. $40m$
• C. $60m$
• D. $80m$

Answer 1

The correct option is D

$80m$

Step 1. Given data:

The initial speed of the automobile, $u_1=60km/h$

The final speed of the automobile, $v_1=0km/h$

The initial speed of the car, $u_2=120km/h$

The final speed of the car, $v_2=0km/h$

The stopping distance of the automobile, $s_1=20m$

Step 2. The formula used:

The third equation of motion, $v^2=u^2+2as.....\left(1\right)$

where $"a"$ is acceleration for car and automobile in $m/s^2$

Step 3. Finding the stopping distance of the car:

The equation of motion for the automobile, $u_1^2=-2a{s}_1.....\left(2\right)$

The equation of motion for the car, $u_2^2=-2a{s}_2.....\left(3\right)$

By dividing equation $\left(3\right)$ by $\left(2\right)$, we get

$$\begin{gathered} {\left(\frac{u_2}{u_1}\right)}^2=\frac{s_2}{s_1} \\ \Rightarrow {\left(\frac{120}{60}\right)}^2=\frac{s_2}{20} \\ \Rightarrow s_2=80m \end{gathered}$$

Hence, the stopping distance of the car is $80m$.

Option D is correct.