# Graphing Kinetic Energy over time

Shaybay92
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.

Gold Member
$$KE=\frac{1}{2}mv^2$$

$$v=at$$

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

$$KE=\frac{ma^2}{2}t^2$$

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)

Shaybay92
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!