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March 25, 2017

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During the first 18 minutes of a 1.0-hour trip, a car has an average speed of 11 m/s. What must the average speed of the car be during the last 42 minutes of the trip be if the car is to have an average speed of 21 m/s for the entire trip?

  • physics - ,

    1hour = 3600s.
    d = 21m/s * 3600s=75600m = t0tal distance to be traveled.

    t1 = 18min.=18min * 60s/min = 1080s.
    d1 = 11m/s * 1080s = 11880m. = Distanced traveled.

    d2 = d - d1,
    d2 = 75600 - 11880 = 63720m = distance remaining.

    t2 = 42min = 42min * 60s/min = 2520s = Time remaining.

    V(avg) = d2 / t2 = 63720 / 2520 = 25.3m/s = Average speed during t2.

    CHECK:

    V(avg) = d / (t1 + t2) = 75600 / 3600 = 21m/s = Average speed for the trip.

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