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Velocity from force/time graph

  1. Oct 25, 2004 #1

    ^^ So according to that graph, force (and so acceleration) increase from t =0 to t=4 then force drops to zero and there's probably a constant deacceleration for two seconds....

    How do I get the velocity at t = 6?

    acceleration is changing as forces changes (increases) so I can't pin it down and Im guessing in order to get velocity at t = 6 I'll need velocity at t = 4 and acceleration between t = 4 and t= 6...

    Any hints/ideas? Thanks for any help. :smile:
  2. jcsd
  3. Oct 25, 2004 #2


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    HINT: If force is proportional to the derivative of the speed then the speed should be proportional to the integral of the force! :-)
  4. Oct 25, 2004 #3
    I alternated between staring at that hint and the graph for 15 minutes...

  5. Oct 25, 2004 #4
    the velocity is the integral of acceleration....acceleration and force are proportional as you know....now you know what an integral represents, right...?
  6. Oct 25, 2004 #5
    ... so velocity is the area under the force? 1/2*b*h = 1/2*4*10 = 20 for initial velocity at t = 4?
  7. Oct 25, 2004 #6
    hah never mind I got it! the hight of the triangle should've been in acceleration values and not force.. so I divided ten by the mass and THEN calculated the area and that's the initial velocity which is equal to the final one at t = 6 because acceleration is 0 from t =4 to t=6.

    Thanks! :biggrin:
  8. Oct 25, 2004 #7
    very good! :)
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