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}}=4{{s}_{1}}=4\times 20=80\,m $