On a two lane road , car (A) is travelling with a speed of `36 km h^(-1)`. Tho car ` 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 ` km, (B) decides to overtake ` A before C does , What minimum accelration of car (B) is required to avoid and accident.

Explanation

Velocity of car ` A= 36 km h^(-1) = 10 ms^(-1)` , Velocity oa cr ` A ot C= 54 km h^(-1) = 15 ms^(-1)` ,
Relative velocity of ` B w.r.t. A =15 ms^(-1)` , Relative velocity of ` C w.r.t. A = 15 + 10 = 25 ms^(-10`
As, ` AB =AC = 1 km = 1000 m`
Time available to ` B or C` for crossing ` A = ( 1000)/(25) = 40 s`
If car ` (B) accelrates with accleration ` a, to cross (A) bofore car (C ) does. then
` u= 5 ms^(-1) , t= 40 s, S= 1000 m , a= ?`
Using , `S = ut + 1/2 at^2 `, we have `1000 = 5xx 40 + 1/2 xx a xx 40^@ or 1000 - 200 = 800 a or a =1 s//s^2`