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Physics Question on Motion in a straight line

On a two-lane road, car AA is travelling with a speed of 36 km h−136 \,km\, h^{-1} . Two cars BB and CC approach car AA in opposite directions with a speed of 54 km h−154\, km \,h^{-1} each. At a certain instant, when the distance ABAB is equal to ACAC , both being 1 km1 \,km , BB decides to overtake AA before CC does. The minimum required acceleration of car BB to avoid an accident is

A

1 m s −21\,m\,s\,^{-2}

B

1.5 m s −21.5\,m\,s\,^{-2}

C

2 m s −22\,m\,s\,^{-2}

D

3 m s −23\,m\,s\,^{-2}

Answer

1 m s −21\,m\,s\,^{-2}

Explanation

Solution

Velocity of car AA , vA=36 km h−1=36×518 m s−1=10 m s−1v_{A}=36\,km\,h^{-1}=36\times\frac{5}{18}\,m\,s^{-1}=10\,m\,s^{-1} Velocity of car BB , vB=54 km h−1=54×518 m s−1=15 m s−1v_{B}=54\,km\,h^{-1}=54\times\frac{5}{18}\,m\,s^{-1}=15\,m\,s^{-1} Velocity of car CC , vC=−54 km h−1v_{C}=-54\,km\,h^{-1} =−54×518 m s−1=−15 m s−1=-54\times\frac{5}{18}\,m\,s^{-1}=-15\,m\,s^{-1} Relative velocity of car BB w.r.t. car AA vBA=vB−vA=15 m s−1−10 m s−1=5 m s−1v_{BA}=v_B-v_A=15\,m\,s^{-1}-10\,m\,s^{-1}=5\,m\,s^{-1} Relative velocity of car CC w.r.t. car AA is vCA=vC−vA=−15 m s−1−10 m s−1=−25 m s−1v_{CA}=v_C-v_A=-15\,m\,s^{-1}-10\,m\,s^{-1}=-25\,m\,s^{-1} At a certain instant, both cars BB and CC are at the same distance from car AA i.e. AB−BC=1 km=1000 mAB - BC = 1 \,km = 1000 \,m Time taken by car CC to cover 1 km1 \,km to reach car AA =1000 m25 m s−1=40 s=\frac{1000\,m}{25\,m\,s^{-1}}=40\,s In order to avoid an accident, the car BB accelerates such that it overtakes car AA in less than 40 s40 \,s . Let the minimum required acceleration be a. Then, u=5 m s−1u = 5 \,m \,s^{-1} , t=40 st = 40 \,s , S=1000 mS = 1000 \,m , a=?a = ? As S=ut+12at2S=ut+\frac{1}{2}at^{2} ∴1000=5×40+12×a×402\therefore 1000=5\times40+\frac{1}{2}\times a\times40^{2} 800a=1000−200=800800a = 1000 - 200 = 800 or a=1 m s−2a = 1 \,m \,s^{-2}