Question

An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, ie., 120 km/h, the stopping distance will be:

• A. 20m
• B. 40m
• C. 60 m
• D. 80 m

Answer 1

Answer: (D)

The braking retardation will remain same and assumed to be constant, let it be $a$ . From 3rd equation of motion, $v^2=u^2+2as$ 1st case: $0={\left(60\times \frac{5}{18}\right)}^2-2a\times s_1$ $\Rightarrow$ $s_1=\frac{{(60\times 5/18)}^2}{2a}$ 2nd case: $0={\left(120\times \frac{5}{18}\right)}^2-2a\times s_2$ $\Rightarrow$ $s_2=\frac{{\left(120\times 5/18\right)}^2}{2a}$ $\therefore$ $\frac{s_1}{s_2}=\frac{1}{4}\Rightarrow s_2=4s_1=4\times 20=80m$