Now I know some of the obvious answers to this, such as if you integrate the acceleration twice, you’ll get time squared, but what I’m really looking for is more of an intuitive answer. One of the other thing that bugs me is the 1/2 in the equation of motion. Why half? Why not 2? And if you say that it relates to conservation of energy, why is only that THAT particular constant satisfies the equation?
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1$\begingroup$ Possible duplicates: physics.stackexchange.com/q/89590/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Nov 2, 2021 at 5:30
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2 Answers
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When an object initially at rest has uniform acceleration $a$ for time $t$, it reaches speed $at$. Its average speed up to time $t$ is $(0+at)/2=\frac12at$. Therefore, the distance travelled is $\frac12at\times t=\frac12at^2$.
If the object has constant mass $m$, a force $ma$ was responsible for this acceleration. (We'll neglect any wasting of energy due to e.g. friction, because we only care about the net force.) Over a distance $\frac12at^2$, this force did work equal to the product of force and distance, i.e. $ma\times\frac12at^2=\frac12ma^2t^2$. In terms of the final speed $v=at$, this work is $\frac12mv^2$. By energy conservation, the kinetic energy has increased from $0$ at rest to this final value of $\frac12mv^2$, so that must be the formula for kinetic energy.
If the ideas above aren't intuitive enough, you may like dimensional analysis. The dimensions of speed and acceleration are respectively $\mathsf{LT^{-1}}$ and $\mathsf{LT^{-2}}$. If we want to get a distance travelled from acceleration, we'll clearly need two factors of time. By Newton's second law force has dimension $\mathsf{LMT^{-2}}$, and work done has dimension $\mathsf{L^2MT^{-2}}$, the same as mass times speed squared. The one thing dimensional analysis can never tell us, however, is that a factor of $\frac12$ comes up. For that, you need something like the average-speed argument above. (It is technically the value of an integral, but it's much easier to work it out from the area of a triangle.)
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$\begingroup$ Great answer! The average acceleration part was the piece I needed to solve the puzzle. The dimensional analysis one makes sense but again, it doesn’t explain the 1/2. But I understood the answer. Thanks! $\endgroup$– AceCommented Nov 2, 2021 at 16:04
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I’m really looking for...more of an intuitive answer.
How about this?
Velocity is the rate of change in position. If we measure position in meters ($m$), then we measure velocity in meters per second ($m/s$). (I.e., how much the position ($m$) changed in each second.)
Acceleration is the rate of change in velocity. If we measure velocity in meters per second, then we measure acceleration in meters per second per second. (I.e., how much the velocity ($m/s$) changed in each second.)
The square comes in from the "...per second per second" part.
$\frac{A m/s}{s} = A m/s^2$
