Two cars, 15 km apart one is turning at a speed of 50kmph other at 40kmph . How will it take to two cars meet.

Two cars, 15 km apart one is turning at a speed of 50kmph
other at 40kmph . How will it take to two cars meet.

Ques:- Two cars, 15 km apart one is turning at a speed of 50kmph other at 40kmph . How will it take to two cars meet.
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2 Answers on this Question

  1. A is travelling at 50kmph, B is travelling at 40
    kmph……..according to the formula
    time taken to meet = distance between them
    ———————-
    relative speed of two vehicles
    so, time taken to meet= 15/(50-40)=15/10=3/2hrs

  2. lets say car A is travelling at 50Kmph and Car B at 40 Kmph.
    Since the question is not very clear, we have two
    possibilities. That the cars are either travelling in same
    direction or they are travelling in opposit directions.
    If the cars are travelling in the same direction then again
    there are two possibilities
    First: Car B is behind Car A – then the cars will never
    meet.
    Second: Car A is behind Car B then they will meet after 90
    mins (one and a half hour) Car B travells 60 Kms in 90 mins
    and Car A travells 75 Kms in 90 mins. Car A is 15 Kms
    behind Car B so it has to cover those 15 Kms in the same
    time to reach car B which is 60+15 = 75 kms.
    If the cars are travelling in the opposite direction then
    they will meet in approximately 10 minutes. this can be
    calculated as follows.
    let us say they meet at a point X as shown below
    A ———————–X—————-B
    —-> 15-X <--- ! ----> X <---- since both travel for the same amount of time let the time taken be T hours.(since the speed is in KMPH time taken will be calculated in hours) From the above B has to travel X Kms and A the remaining distance which is 15-X kms ( Distance between A & B is 15Kms) From the formula Distance = Speed * Time we can say Time = Distance / Speed Hence T(A) = (15-X)/50 _______ Eq 1 and T(B) = X/40 __________ Eq 2 Since T(A) = T(B), we have (15-X)/50 = X/40 ______Eq 3 By solving the above Eq 3 we get X = 6.666 Kms Substituting X in either of Eq 1 or Eq 2 we get T = 0.1666 Hrs which is approximately 10 Mins.

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