can you provide a shortcut for the following sum?
- on a 2-lane road, car A is travelling with a speed of 36kmph. Two cars B & C approach A in opposite directon with speed of both being 54kmph. At a certain instant when distance AB=AC,both beng 1 km,B decides to overtake A before C does.The minimum required acceleration of car B to avoid an accident is?
Asked by imtiyazmulla68
4th March 2018,
9:32 PM
Answered by Expert
Answer:
Speed of A, 36 km/hr = 10 m/s
Speed of B and C , 54 km/hr = 15 m/s
Let time taken by B to cross A before C with minimum acceleration is t seconds.
A travel 10×t m in t seconds.
If B crosses A with v m/s then the distance travelled by B in t seconds is ( (v+15)/2 )×t ; we have assumed average velocity of B is (v+15)/2 m/s.
Since the total distance travelled by A and B is 1 km, we write
10×t + ( (v+15)/2 )×t = 1000
let us eliminat t using t = (v-15)/a, where a is the minimum acceleration of B;
Answered by Expert
7th March 2018,
12:07 AM
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