Timeline for Deriving the equation for kinetic energy
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 2, 2018 at 7:04 | history | edited | J.G. | CC BY-SA 4.0 |
added 12 characters in body
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| Jul 12, 2017 at 12:32 | vote | accept | Pancake_Senpai | ||
| Jul 12, 2017 at 12:32 | comment | added | Pancake_Senpai | Oh right! That was a really silly mistake. Thank you. | |
| Jul 12, 2017 at 12:26 | comment | added | Alex S | It is a big difference between the equation of motion of a mass going with constant velocity and the equation of motion of a mass going with constant acceleration. The first equation is $\Delta d=v*\Delta t$ and the last one is $\Delta d=va*\Delta t$. Again, you applied the first one, not the second. You perhaps don't realize that because $vi=0$ | |
| Jul 12, 2017 at 12:20 | comment | added | Alex S | This equation of motion is not correct. When a mass goes with a constant acceleration, the correct equation of motion is $\Delta d=va*\Delta t$, where $\Delta d$ is the difference $xf-xi$ corresponding with the final and initial coordinates of the mass, va is the average velocity $va=(vi+vf)/2$ and $\Delta t$ is the time. In your case, $va$ is not $vf$ is $vf/2$ or according to your notation is $v/2$ and you got the correct solution with $2/v$ | |
| Jul 12, 2017 at 12:11 | comment | added | Pancake_Senpai | $\frac{\Delta t}{\Delta d} = \frac{1}{\Delta v}$, but when the object starts from rest then $\Delta v = $ final velocity of particle, which I'm representing by $v$. | |
| Jul 12, 2017 at 11:53 | comment | added | Alex S | Then you made a mistake when you said that $\Delta t/\Delta d=1/v$. This assumption would have been correct if the mass would have been going with constant velocity, but is not the case here | |
| Jul 12, 2017 at 11:47 | review | First posts | |||
| Jul 12, 2017 at 12:09 | |||||
| Jul 12, 2017 at 11:46 | comment | added | Pancake_Senpai | In regards to both your answer and the other answer by @Steven , I understand your answers but I don't understand what I did wrong in my derivation. I must have made an error otherwise I would have had my missing $\frac{1}{2}$. | |
| Jul 12, 2017 at 11:43 | history | answered | Alex S | CC BY-SA 3.0 |