Graphing Kinetic Energy over time

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SUMMARY

The discussion centers on calculating kinetic energy (KE) over time for an object in freefall or descending a frictionless hill. The correct formula for kinetic energy is KE = 1/2 mv^2, where v is derived from the equation of motion v = at. The author initially attempted to differentiate the kinetic energy equation directly, leading to incorrect cubic results. The correct relationship established is KE = (ma^2/2)t^2, indicating that kinetic energy increases quadratically with time due to constant acceleration.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with kinetic energy equations
  • Basic calculus, specifically differentiation
  • Knowledge of freefall dynamics and gravitational acceleration
NEXT STEPS
  • Study the derivation of kinetic energy from the equations of motion
  • Learn about the implications of constant acceleration on motion
  • Explore graphing techniques for visualizing kinetic energy over time
  • Investigate the effects of friction on kinetic energy calculations
USEFUL FOR

Students and educators in physics, engineers working on motion dynamics, and anyone interested in understanding the relationship between kinetic energy and time in freefall scenarios.

Shaybay92
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If an object is in freefall (or in my case traveling 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 doesn't 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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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)
 
Oh my goodness I can't believe 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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