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 ?
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Answer 1

Solution :-

As per the given data ,

• Speed of  car A = 36 km / hr = 36 x 5 / 18 = 10 m /s

• Speed of Car B = 54 km / hr = 54 x 5 / 18 = 15 m /s

• Speed of Car C = - 54 km / hr = - 15 m /s

Speed of Car B with respect to A (Vba)

= Vb - Va

= 15 - 10  

= 5 km / hr

Speed of car C with respect to A (Vca)

= Vc - Va

= -15- 10

= - 25 km / hr

Time required by the C to overtake car A

we know that ,

⇒ speed = distance / time

⇒ time = distance / speed

hence ,

⇒ t = 1000 / 25

Now as per the question , B needs to overtake A before C does in order to do that he has maximum 40 s

Now we need to find the minimum acceleration of the car B required to avoid accident ,

By using second equation of motion ,

⇒ S = Vba x t + a t² / 2

⇒ 1000 = 5 x 40 + a x 1600 /2

⇒ 1000 = 200 + a x 800

⇒ 800= 800 a

⇒ a = 1 m /s ²

The maximum acceleration of the car in order to avoid accident is 1 m /s²