Question

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?

Answer 1

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

= 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 = 40 s

Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.

From the second equation of motion, the minimum acceleration (a) produced by car B can be obtained as:

`s = ut + 1/2 at^2`

`1000 = 5 xx 40 +1/2 xx a xx (40)^2`

 `a= 1600/1600 = 1 "m/s"^2`

Answer 2

Speed of car A, `v_A = 35 "km/h" = 36 xx 5/18 = 10 "m/s"`

Speed of car B, `v_B = 54 "km/h" = 54 xx 5/18 = 15 "m/s"`

Relative speed of car A w.r.t car `C = v_(AC) = (10+15) ms^(-1) = 25 ms^(-1)`

Relative speed of car B w.r.t car `A = v_(BA) = (15 - 10) ms^(-1) = 5 "m/s"`

Time taken by car C to cover distance AC,

`t = 1000/v_(AC) = 1000/25 = 40s`

if a is the acceleration, then

`S = ut + 1/2 at^2`

`=>1000 =  5 xx 40 +1/2a xx (40)^2`

`=> a = (1000 - 200)/800` = 1 m/s²