Graphing Kinetic Energy over time

  • Thread starter Shaybay92
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  • #1
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If an object is in freefall (or in my case travelling down a hill without friction) then what function could you use to calculate the kinetic energy gain over time?

I have attempted to use equations of motion and substitute them into the kinetic energy equation:

KE = 1/2 mv^2
KE = 1/2 m(4.905t^2)^2
and then differentiate with respect to t in order to have a function that describes kinetic energy over time. This doesnt seem to work. I actually differentiated my v = ut + 1/2at^2 before I put it into the KE equation and got the right answer, but differentiating BEFORE I put it into the kinetic energy equation seems wrong? I want kinetic energy over time so therefore I should differentiate the KE equation with respect to t. This gives me:

KE = 1/2 m(4.905^2)*t^3 which is obviously incorrect because kinetic energy is not represented by a cubic function, and gives me entirely wrong answers. It doesn't make sense.
 

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  • #2
Matterwave
Science Advisor
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[tex]KE=\frac{1}{2}mv^2[/tex]

[tex]v=at[/tex]

V is NOT .5at^2, that's displacement (or y).
Therefore:

[tex]KE=\frac{ma^2}{2}t^2[/tex]

That would be the relation between kinetic energy and time. As time goes by, v increases linearly as t since the acceleration is constant. Therefore, the kinetic energy increases quadratically in t since it increases quadratically in v.

This is only valid for before the object hits the ground of course (and for distances not too far above the Earth's surface)
 
  • #3
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Oh my goodness I can't beleive I made such a silly mistake as 'not remembering my equations of motion'. The whole point of me doing this graph was to prove that I know how to use the equations!!!! Thankyou so much for pointing out how oblivious I was, and consequently saving me from several hours of additional frustration!
 

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