Question

On a two-lane road, car $ A $ is travelling with a speed of $ 36 \,km\, h^{-1} $ . Two cars $ B $ and $ C $ approach car $ A $ in opposite directions with a speed of $ 54\, km \,h^{-1} $ each. At a certain instant, when the distance $ AB $ is equal to $ AC $ , both being $ 1 \,km $ , $ B $ decides to overtake $ A $ before $ C $ does. The minimum required acceleration of car $ B $ to avoid an accident is

• A. $ 1\,m\,s\,^{-2} $
• B. $ 1.5\,m\,s\,^{-2} $
• C. $ 2\,m\,s\,^{-2} $
• D. $ 3\,m\,s\,^{-2} $

Answer 1

The Correct Option is A

Velocity of car $ A $ , $ v_{A}=36\,km\,h^{-1}=36\times\frac{5}{18}\,m\,s^{-1}=10\,m\,s^{-1} $ Velocity of car $ B $ , $ v_{B}=54\,km\,h^{-1}=54\times\frac{5}{18}\,m\,s^{-1}=15\,m\,s^{-1} $ Velocity of car $ C $ , $ v_{C}=-54\,km\,h^{-1} $ $ =-54\times\frac{5}{18}\,m\,s^{-1}=-15\,m\,s^{-1} $ Relative velocity of car $ B $ w.r.t. car $ A $ $ v_{BA}=v_B-v_A=15\,m\,s^{-1}-10\,m\,s^{-1}=5\,m\,s^{-1} $ Relative velocity of car $ C $ w.r.t. car $ A $ is $ v_{CA}=v_C-v_A=-15\,m\,s^{-1}-10\,m\,s^{-1}=-25\,m\,s^{-1} $ At a certain instant, both cars $ B $ and $ C $ are at the same distance from car $ A $ i.e. $ AB - BC = 1 \,km = 1000 \,m $ Time taken by car $ C $ to cover $ 1 \,km $ to reach car $ A $ $ =\frac{1000\,m}{25\,m\,s^{-1}}=40\,s $ In order to avoid an accident, the car $ B $ accelerates such that it overtakes car $ A $ in less than $ 40 \,s $ . Let the minimum required acceleration be a. Then, $ u = 5 \,m \,s^{-1} $ , $ t = 40 \,s $ , $ S = 1000 \,m $ , $ a = ? $ As $ S=ut+\frac{1}{2}at^{2} $ $ \therefore 1000=5\times40+\frac{1}{2}\times a\times40^{2} $ $ 800a = 1000 - 200 = 800 $ or $ a = 1 \,m \,s^{-2} $