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 ` 1 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, `v_A` = 36 km/h = 10 m/s
Velocity of car B, `v_B` = 54 km/h = 15 m/s
Velocity of car C, `v_C` = 54 km/h = 15 m/s

Relative velocity of car B with respect to car A,
`v_(BA) = v_B – v_A` ltbrge 15 – 10 = 5 m/s
Relative velocity of car C with respect to car A,
`v_(CA) = v_C – (– v_A)`
= 15 + 10 = 25 m/s
At a certain instance, both cars B and C are at the same distance from car A i.e.,s = 1 km = 1000 m

Time taken (t) by car C to cover 1000 m =`1000/25=40s`
Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.
From second equation of motion, minimum acceleration (a) produced by car B can be obtained as:
`s=ut+1/2at^2`
`1000=5xx40+1/2xxaxx(40)^2`
`a=1600/1600=1 m//s^2`