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Kinematics-Based Question

  1. Oct 7, 2006 #1
    If a trip takes 2.8 h at an average speed of 100 km/h, how much time will be saved if the speed is increased by 10km/h?

    So this is what I tried to do:

    2.8 = x
    100 110

    100x = 2.8(110)

    100x = 308

    x = 3.08

    3.08 h - 2.8 h
    = 0.28 h

    I came to the conclusion that the amount of time that will be saved is 0.28 h.

    Can anyone tell me if this is right or if there's another way of doing this, for example using a Kinematics equation?
  2. jcsd
  3. Oct 7, 2006 #2
    you can use the kinematics equation...

    Xf = Xi + ViT + 1/2at^2

    No acceleration in this problem so a = 0. The initial position (Xi) = 0 so...

    Xf = ViT

    Xf = (100 km/h)(2.8h)
    Xf = 280km

    So the trip is 280km and takes 2.8 hours, in the second part they want time as the variable and want use to use the trip length as a constant.

    (280km) = (110 km/h)(T)
    T = 2.545 hours

    2.8 - 2.545 = .255 hours
    Last edited: Oct 7, 2006
  4. Oct 7, 2006 #3


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    Science Advisor
    Homework Helper
    Gold Member

    No, that is not correct. First calculate the distance travelled in 2.8 hours at 100km/hr; then calculate the time it would take to travel that distance at the increased average speed of 110 km/hr; then determine the savings in time.
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