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On a two-lane road, car A is travelling with a speed of 36 km h-1. Two cars B andC approach car A in opposite directions with a speed of 54 km h-1 each. At acertain instant, when the distance AB is equal to AC, both being 1 km, B decidesto overtake A before C does. What minimum acceleration of car B is required toavoid an accident ? |
Answer» Solution :-As per the given data ,
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 |
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