IIT JEE , AIEEE Forum - Physics , Mechanics - a car travelling at 60km/h overtakes another car travelling at 42km/h.Assuming each car to

elessar_iitkgp

Let the time of overtake be T. The overtake starts at the moment when the second car's front is just behind the first car's rear. The overtake ends when the first car's rear is in front of the second car's front. The distances moved by the cars are 60T and 42T
It can be seen by drawing a diagram that
60T - 42T = 10/1000
T = 1/1800 hrs = (1/1800)(3600) s = 2 s

The total road distance used for the overtake is = 60T + 5/1000 km = 60(1/1800) + 5/1000 km = 23/600 km = (23/600)(1000)

38 m.
You are all committing the same mistake ... the total road distance is asked and not the distance moved by the second car ... think carefully and try to understand the difference between them. If I remember correctly, this is a problem from HCV exercise ...

The road distance is the distance moved by the second car + the length of the second car itself. The question asks for the length of the road utilized and not the distance that the second car moves to overtake the first one.

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