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Thread: Rate of Change - Acceleration

  1. #1
    Jul 2009

    Rate of Change - Acceleration

    A car is traveling C meters behind a truck, both traveling at a constant speed of V m/s. The road widens L meters ahead of the truck and there is an overtaking lane. The car accelerates at a uniform rate so that it is exactly alongside the truck at the beginning of the overtaking lane.

    a)What is the acceleration of the car?
    b) Show that the speed of the car as it passes the truck is V(1+ 2C/L)

    Have tried part a), but gotten (V^2C)/(L^2) instead it should be (2V^2C)/(L^2)

    Any thoughts?

    This question is in my book as Rates of Change, but i am not sure as my approach was non - calculus based.
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  2. #2
    MHF Contributor

    Apr 2005
    It's impossible to tell what you did wrong if you don't show exactly what you did. HOW did you get "(V^2C/L^2)" as the acceleration?

    The truck is traveling at constant velocity V so if we let "t" be the time until the truck reaches the passing lane, Vt= L (and so t= L/V).

    With constant acceleration, a, the car travels distance (1/2)at^2+ Vt in time t and, because it is distance C behind the truck, must travel distance L+ C in the same time to be beside the truck: (1/2)at^2+ Vt= L+ C= Vt+ C. We can cancel the "Vt" terms to get (1/2)at^2= C. Put t= L/V into that. I don't see any need for Calculus here.
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