Solveeit Logo
Login

Question

Question: On a two-lane road, car A is travelling with a speed of 36 km h<sup>-1</sup>. Two cars B and c appro...

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. The minimum required acceleration of car B to avoid an accident is

A

1 m s-2

B

1.5 m s-2

C

2 m s-2

D

3 m s-2

Answer

1 m s-2

Explanation

Solution

Velocity of car A,

VA=36kmh−1=36×518ms−1=10ms−1V_{A} = 36kmh^{- 1} = 36 \times \frac{5}{18}ms^{- 1} = 10ms^{- 1}

Velocity of car B,

VB=54kmh−1=54×518ms−1=15ms−1V_{B} = 54kmh^{- 1} = 54 \times \frac{5}{18}ms^{- 1} = 15ms^{- 1}

Velocity of car C,

VC=−54kmh−1=−54×518ms−1=−15ms−1V_{C} = - 54kmh^{- 1} = - 54 \times \frac{5}{18}ms^{- 1} = - 15ms^{- 1}

Relative velocity of car B w.r.t. car A

VBA=VB−VA=15ms−1−10ms−1=5ms−1V_{BA} = V_{B} - V_{A} = 15ms^{- 1} - 10ms^{- 1} = 5ms^{- 1}

Relative velocity of car C w.r.t. car A is

VCA=VC−V⥂A=−15ms−1−10ms−1=−25ms−1V_{CA} = V_{C} - V ⥂_{A} = - 15ms^{- 1} - 10ms^{- 1} = - 25ms^{- 1}

At a certain instant, both cars B and C are at the same distance form car A

i.e., AB = BC = 1 lm = 1000m

The taken by car C to cover 1 km to reach car a =1000m25ms−1=40s= \frac{1000m}{25ms^{- 1}} = 40s

In order to avoid an accident, the car B accelerates such that it overtakes car A in less than 40s. Let the minimum required accelerations be a. Then,

u=5ms−1u = 5ms^{- 1}, t = 40s, S = 1000 m, a = ?

As S=ut+12at2S = ut + \frac{1}{2}at^{2}

∴1000=5×40+12×a×402\therefore 1000 = 5 \times 40 + \frac{1}{2} \times a \times 40^{2}

800a = 1000 – 200 = 800 or a = 1 ms−2ms^{- 2}