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Posted by on Tuesday, August 9, 2011 at 12:13am.

An automobile moving at a constant velocity of 15m/s passes a gasoline station. Two seconds later, another automobile leaves the gasoline station and accelerates at a constant rate of 2m/s^2 in the same direction. How soon does the second autombile overtake the first?

  • Physics - , Tuesday, August 9, 2011 at 6:28am

    Let t =0 be the time after the first can passes by.

    Solve for t when
    15 t = (1/2)*(2.0)*(t-2)^2

    The term on the right is the distance covered by the accelerating car that starts 2 seconds late.

    15t = (t-2)^2 = t^2 -4t + 4
    t^2 -19t +4 = 0

    t = [19 +sqrt(345)]/2 = 18.8 seconds

    They both will be 282 m from the gas station at that time.

  • Physics - , Thursday, January 31, 2013 at 10:24am

    asjklaj;a

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