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March 26, 2017

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Cars A and B are racing each other along a straight path in following manner: Car A has head start and is a distance dA beyond starting line at t=0. The starting line is at x=o. Car A travels at constant speed vA. Car B starts at starting line but has better engine than car A and travels at constant speed vB (which is greater than vA). How long after Car B started the race will Car B catch up with Car A? How far from Car B's starting line will the cars be when Car B passes Car A?

  • physics - ,

    To summarize:
    At t=0
    car A: x=dA, speed=vA
    car B: x=0, speed=vB
    When will car B overtake car A and where.

    Distance to catch-up = dA
    difference in speed = (vB-vA)
    Time to catch up, T = dA/(vB-vA)
    location where the two cars are side-by-side
    = T*vB
    = dA*vB/(vB-vA)

  • physics - ,

    sdjg

  • physics - ,

    Da/(Vb-Va)
    Is the right answer.

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