Chuck’s car is moving at 65.0m/s when he suddenly accelerates his car at 15.0m/s2 for 3.00s. How far did Chuck, and his car, travel while he was accelerating?


  1. Answer:

    x = 265.5 m


    To find the distance traveled by the car you first use the following formula:

    v=v_o+at   (1)

    vo: initial velocity = 65.0m/s

    a: acceleration = 15.0m/s^2

    t: time = 3.00 s

    you replace the values of vo, t and a in the equation (1) in order to calculate the final velocity v:

    v=65.0m/s+15.0m/s^2(3.00s) = 110 m/s

    Next, you use the following formula:


    x: distance traveled

    You do x the subject of the formula and replace the values of vo, v and a:

    x=\frac{v^2-v_o^2}{2a}\\\\x=\frac{(110m/s)^2-(65.0m/s)^2}{2(15.0m/s^2)}=262.5\ m

    hence, the distance traveled by the car is 265.5 m


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