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Posted by on Sunday, November 30, 2008 at 7:27pm.

A car is traveling along a straight road at a velocity of +31.0 m/s when its engine cuts out. For the next ten seconds, the car slows down further, and its average acceleration is a1. For the next five seconds, the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the fifteen-second period is +24.5 m/s. The ratio of the average acceleration values is a1/a2 = 1.67. Find the velocity of the car at the end of the initial ten-second interval.

  • Physics - , Sunday, November 30, 2008 at 7:48pm

    OK, so a1 = 1.67 a2
    After 15 seconds, the velocity is
    31 + a1*10 + a2*5
    = 31 + 1.67 a2 *10 + a2* 5
    = 31 + 21.7 a2 = 24.5
    21.7 a2 = -6.5
    a2 = -0.30 m/s^2
    a1 = -0.50 m/s^2
    V(@ 10 s) = V + a1*10 = ?

  • Physics - , Sunday, November 30, 2008 at 8:54pm

    31+ (-.5)(10)

    26

    Is that right?

  • Physics - , Sunday, November 30, 2008 at 9:49pm

    Yes

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