On a two-lane road, 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 = 36 km `h^(-1)=10 ms^(-1)`
Velocity of car B or C = 54 km `h^(-1)=15ms^(-1)`
Relative velocity of B w.r.to
A = 15 - 10 = 5 `ms^(-1)`
Relative velocity of C w.r.to
A = 15 + 10 = 25 `ms^(-1)`
As, AB = AC = 1 km = 1000 m
Time available to B (Or) C for crossing
`A=(1000)/25=40s`
If car B accelerates with acceleration a, to cross A before car C does, then
`" "u=5ms^(-1), t=40s, s=1000m, a=?`
Using `s=ut+1/2at^(2)`
`1000=5 times 40 +1/2 times a times 40^(2)` (Or)
1000 - 200 = 800 a (Or)
a = 1 `m//s^(2)`