- #1
Shaybay92
- 124
- 0
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.
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.