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Homework Help: Force vs Time graph (Impulse)

  1. Jan 21, 2013 #1
    1. The problem
    I have trouble understanding slope and area on graphs. For a force as a function of time graph the area is equal to the impulse. Why is that? And what is the slope equal to? How do I know?

    2. Relevant equations

    3. The attempt at a solution
    So the slope should be equal to f/t?
    What on earth is force divided by time equal to?
    And is the area underneath the graph always equal to y*x (in this case y=force, x=time)? So area = f*t = impulse?
  2. jcsd
  3. Jan 21, 2013 #2


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    Homework Helper

    The impulse of a force is defined as I=FΔt for a force F acting for a very short time. And the change of momentum p=mv is equal to the impulse of the force. I=Δp=mvf-mvi.

    See the first picture: a constant force is plotted. Its impulse is FΔt= F(tf-ti), the area under the line F(t) between the initial and final times.

    The second picture shows a force which grows linearly with time. Its impulse is equal to the area under the line, which is the same as the average force multiplied by the elapsed time: I=FavΔt=Fav(tf-ti).
    The picture on the right shows a general force-time plot. Again, the impulse of the force between ti and tf is equal to the area under the curved line.

    If you know the impulse of a force imparted to a body, it is equal to the change of momentum. I=mvf-mvi.

    "f " means the average force during the time period Δt.

    The slope of the f(t) graph is irrelevant. It is not f/t, but the limit of Δf/Δt when Δt get shorter and shorter. (If you study Calculus you will learn that it is the time derivative of f(t).)

    The area is fΔt in case f is constant. Otherwise you slice the whole area and add the small pieces f(t)Δt together. You will learn that the area is equal to the integral of f(t) between ti and tf.


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