University of Wisconsin Green Bay

According to the website, a 2010 Ford Mustang GT can acceleration from zero to 60 mph ( 26.8 m/s) in 4.9 s. What is the average acceleration of the car during that period of time? If acceleration on the car is constant, how far does the car travel in those 4.9 s?

  • In this problem, you are asked to describe the motion (acceleration, location) of the car. Whenever you are asked to describe the motion of an object without worrying about the cause of that motion, you have a kinematics problem.

  • There are three key kinematic equations. If you carefully select the equation which most directly describes the situation in your problem, you will not only solve the problem in fewer steps but also understand it better. The three equations, written for motion in the x-direction, are:

    1. x = x0 + v0Δt + ½ a(Δt)2 (relates position and time)
    2. v = v0 + aΔt (relates velocity and time)
    3. v2 = v02 + 2a(Δx) (relates velocity and position)

    a) The acceleration of the car is a variable in all three equations, so this question drives home the point that you can’t pick your equation just by what is requested in the problem. In this case, you want to find the acceleration that is needed for the car to reach a velocity of 26.8 m/s in a time interval of 4.9 s. The question relates velocity and time, so you want to use equation 2.

    b) You are asked how far (position) the car travels in 4.9 s (time), so you want to use equation 1.

  • Step 1:

    The average acceleration of the car as it moves from Point 1 to Point 2 is the information that was requested in the first of the problem. Scroll down to step 2 to answer the second question.


    Step 2:

    The second part of the problem asked for the distance the car traveled in the 4.9 s. No further mathematical solution is required.

  • v2 = v1 + aΔt
    26.8 m/s = 0 + a(4.9 s)
    5.5 m/s2 = a

    In this problem, you were first asked to find the acceleration of the car for the case where it goes from a velocity of 0 to 60 mph in a time of 4.9 s. The answer is a little bit more than half that of gravity (about 0.56g) which is a reasonable answer for the maximum acceleration of a Mustang.

    x2 = x1 + v1Δt + ½ a(Δt)2
    x2 = 0 + (0) Δt + ½ (5.5 m/s2)(4.9 s)2
    x2 = 66 m

    In the second part, you were asked to find how far the car traveled in that amount of time, and so the position-time equation gives the answer most directly.