Example Problems for Kinematics. I

Last revised 1999/09/09


  1. A car accelerates from rest at an acceleration of A for a length of time T. How far does the car travel? (Call the distance travelled X.)

  2. A car decelerates at a rate of -A and comes to rest in a time T. How far (X) does the car travel? Notice that this problem is just what we would see if we videotaped the car in problem 1 and played the tape backwards. The process of running the videotape backwards is called time reversal. Usually physics is unchanged by time reversal, so we should expect to get an answer for problem 2 that is the same as the answer to problem 1.

    Solution for Problem 1:
    We have a problem in one-dimensional motion with constant acceleration, for which the equations are

    x = x0 + v0 t + (1/2) a t2

    v = v0 + a t

    If we choose t=0 at the beginning of the car's motion, then the information we have available is

    x0 = 0
    v0 = 0
    x(T) = X

    and the working equations become

    (1): X = 0 + 0 T + (1/2) A T2
    (2): v = 0 + A T

    Equation (1) can be immediately solved to get X = (1/2) A T 2 and we never need equation (2).


    Solution for Problem 2
    Solution 1:
    Take t=0 at the beginning of the acceleration, as in problem 1. Then we have

    x0 = 0
    v0 = unknown
    v(T) = 0 [because the car comes to a stop]
    x(T) = X

    This time we must use the velocity equation to get v0:

    v = v0 + a t
    0 = v0 - A T
    v0 = A T

    after which we can use the distance equation:

    x = x0 + v0 t + (1/2) a t2
    x = 0 T + (A T) T - (1/2) A T 2
    x = (1/2) A T 2

    Interestingly, we get the same answer as in problem 1. A car stopping from a set velocity travels exactly the same distance as a car accelerating at the same rate and reaching the same velocity. Can we get this answer with less algebra?

    Solution 2:
    Take t=0 at the end of the deceleration instead. When we do this, it is most natural to put x=0 at that point, too. We could, but since we have just worked the problem with x=0 at the beginning we will stick with that choice. Then we have

    v0 = 0
    x0 = X
    x(-T) = 0

    Then the distance equation gives

    0 = X + 0 T - (1/2) A T2
    X = (1/2) A T2

    We get the same result but only have to use one equation.

    In general, it is a good guess to set t=0 at the spot where you have the most variables which are either known or wanted.