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 acceleration of car (B) is required to avoid and accident.

Options

A. `1 m s^(-2)`
B. `1.5 m s^(-2)`
C. `2 m s^(-2)`
D. `3 m s^(-2)`

Answer

A

Explanation

Velocity of car A,
`v_A = 36 km h^(-1) = 36 xx 5/18 m s^(-1) = 10 m s^(-1)`
Velocity of car B, `v_B = 54 km h^(-1) = 54 xx 5/18 m s^(-1) = 15 m s^(-1) `
Velocity of car C,
`v_c = -54 km h^(-1) = -54 xx 5/18 m s^(-1) = -15 m s^(-1)`
Relative velocity of car B w.r.t car A `v_BA = v_B - v_A = 15 m s^(-1) - 10 m s^(-1) = 5 m s^(-1)`
Relative velocity of car c, w.r.t car A is
`v_(CA) = v_C - v_A = -15 m s^(-1) - 10 m s^(-1) = -25 m s^(-1)`
At a certain instant, both cars B and C are at the same distance from car A
i.e. AB = BC = 1 km = 1000 m
Time taken by car C to cover 1 km to reach car A = `1000 m //25 m s^(-1)= 40` s
In order to avoid an accident, the car B accelerates such that it overtakes car A in less than 40 s. Let the minimum required acceleration be a. Then,
`u = 5 m s^(-1), t = 40 s, S = 1000 m, a = ?`
As `S = ut + 1/2 at^(2)`
`therefore 1000 = 5 xx 40 + 1/2 xx a xx 40^(2)`
800a = 1000 - 200 = 800 or a = `1 m s^(-2)`