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?

Explanation

Velocity of car A, `v_A =36 km *h^(-1) = 10 m*s^(-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 B relative to A,
`v_(BA)=v_B-v_A =15-10=5 m*s^(-1)`
Velocity of car C relative to A ,
`v_(CA)=v_C-(-v_A)=v_C+v_A=15+10=25 m*s^(-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)`