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.

Options

A. `9 . 8 ms ^(-2)`
B. ` 10 ms ^(-2)`
C. `1 ms ^(-2)`
D. `2.0 ms ^(-2)`

Answer

C

Explanation

Velocity of car `A,v_A=36 " km h"^(-1) = 36 xx (5)/(18) ms^(-1) = 10 ms ^(-1)`
Velocity of car ` B,v_B=54 " km h"^(-1) = 54 xx (5)/(18) ms^(-1) = 15 ms ^(-1)`
Velocity of car `C ,v_C=54 " km h"^(-1) = 54 xx (5)/(18) ms^(-1) = -15 ms ^(-1)`
Relative velocity of car B w.r.t car A
Relative velocity of car C w.r.t. car A
`v_("BA")=v_b-v_A =15 ms^(-1) -10ms^(-1) =5ms^(-1)`
Relative velocity of car C w.r.t car A is `v_("CA")=v_C-v_A =-15 ms^(-1) -10ms^(-1) =-25ms^(-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
`= (1000m)/(25 ms^(-1))=40 s`
` u = 5 ms^(-1) , t = 40 s, S = 1000 m , a = ? `
As `S = ut +1/2 "at"^(2) , :. 1000 = 5 xx 40 +1/2 xx a xx 40^(2)`
or ` 800 a = 1000 - 200 = 800 " or " a = 1 ms^(-2)`