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

Hii dear,

# Answer- 1 m/s^2

# Explaination-
We have car A and car B going in same direction and car C in opposite direction.
Suppose v1, v2 and v3 be velocities of car A, car B and car C respectively.

- Given is-
v1 = 36 km/h = 10 m/s
v2 = 54 km/h = 15 m/s
v3 = 54 km/h = 15 m/s

Relative velocity of car B with respect to car A,
v21 = v2-v1 = 15–10 = 5 m/s
Relative velocity of car C with respect to car A,
v31 = v3 – (– v1)
= 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 is 1000/25 = 40 s.

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

From second kinematic equation of motion,
s = ut + (1/2)at^2
1000 = 5×40+(1/2)×a×(40)^2
a = 1600/1600 = 1 m/s^2.

Hence, minimum acceleration produced by car B should be 1 m/s^2.

Hope this was helpful...

Answer 2

Explanation:

Velocity of car A, vA = 36 km/h = 10 m/s

Velocity of car B, vB = 54 km/h = 15 m/s

Velocity of car C, vC = 54 km/h = 15 m/s

Relative velocity of car B with respect to car A,

vBA = vB – vA

= 15 – 10 = 5 m/s

Relative velocity of car C with respect to car A,

vCA = vC – (– vA)

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

s = ut + (1/2)at2

1000 = 5 × 40 + (1/2) × a × (40)2

a = 1600 / 1600 = 1 ms-2