On a two-lane 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. What minimum acceleration of car B is required to avoid an accident?

On a two-lane car A is travelling with a speed of 36 "km.h"^(-1) . Two

1.	On a two-lane 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. What minimum acceleration of car B is required to avoid an accident?
Answer» Solution :Velocityof car A, `v_A =36 km h^(-1) = 10 ms^(-1)` [Fig.2.92] . Let `v_B and v_C` be the velocities of CARS B and C. `therefore v_B =v_C =54km h^(-1) =5 m s^(-1)` Velocity of car Brelative to A, `v_(BA)=v_B-v_A =15-10=5 ms^(-1)` Velocity of car C relative to A , `v_(CA)=v_C-(-v_A)=v_C+v_A=15+10=25 ms^(-1)` Also the distance of B and C from A is 1 km =100m. Let t=time taken by car C to travel a distance 1 km towards A . `therefore S=v_(CA)t or, 1000 =25t or, t=40s ` SUPPOSE a =minimumu ACCELERATION required for car B to avoid the accident . With this acceleration , car B overtakes car A in 40 s . So, `S=v_(BA)t+1/2at^2 or, 1000 =5xx40 +1/2axx(40)^2` `therefore a=((1000-200)xx2)/((40)^2)= 1m *s^(-2)`