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Let's say $g=10\ \text{m/s}^2$ and it drops for $3$ seconds. For the first second it will drop $10\ \text{m}$, the second it will drop $20\ \text{m}$ ($10 + 10$) and the third second it will drop $30\ \text{m}$. Then if we add $10+20+30$ it equals $60$ but if you use the equation it equals $45$.

Whats the flaw in my thinking?

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Your flaw lies in determining the number of meters the object falls in each second. Take just the first 1-second interval as an example. You stated it would fall 10 meters, while it will actually fall $gt^2/2=5\ \text{m}$.

The basis of the flaw above is in your interpretation for $g$. This quantity tells you how much the velocity changes every second, not the distance it falls. (You may want to review your textbook for interpretations of acceleration.)

Alright, you asked for the intuition behind $d=\frac{1}{2}gt^2$. To help with that, here is the same equation written in a useful form:

$$d=\frac{0+gt}{2}t$$

The fraction $\frac{0+gt}{2}$ is just the average speed throughout the fall. (This works because the speed increases linearly, meaning $g$ is constant.) The intuition behind your equation, then, is that you're multiplying the average speed by the time to get the distance.

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g is telling you the constant rate at which the velocity is changing. So initially the velocity is 0 and after 1 second the velocity is 10 m/s. The average velocity during that first second is 5 m/sec so the mass has fallen 5 m. In the second second the initial velocity is 10 m/s and at the end of that second it is 20 m/s. The average velocity over that second is 15 m/sec and the mass has fallen an additional 15 m. Finally in the third second the initial velocity is 20 m/s, final velocity is 30 m/s, average is 25 m/s, and thus distance fallen is 25 m. Add the three distances, 5+15+25=45, same as the formula.

Shortcut: Initial velocity is 0, after 3 seconds the final velocity is 30 m/s, so average velocity over 3 second fall is 15 m/s. Distance fallen is 3 seconds times average velocity = 45 m.

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  • $\begingroup$ Thanks that's a very good explanation. what if its velocity increases by the square? how would you calculate distance traveled then? $\endgroup$ – Ray Kay Aug 21 '14 at 5:54
  • $\begingroup$ Well, if there is a simple way to calculate that the average velocity in that case is 1/3 the final velocity(assuming initial velocity is 0) without using calculus, I don't see how at the moment. See the first integral in Bernhard's answer. That's the general case. $\endgroup$ – Robert Miller Aug 22 '14 at 5:24
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You are not making a fundamental error, and your approach is in principle correct. Basically, what you get, is a discretization error. Basically, what you are doing is evaluating the integral

$$d=\int_{t_0}^{t_1}v(t)dt=\int_{t_0}^{t_1} gt dt.$$

Which you approximated with a Riemann sum (maybe unintentionally?), i.e

$$d=\sum_{i=1}^N g t_i \Delta t.$$

You take $N=3$, $\Delta t=(t_1-t_0)/N=1\text{ s}$ and $t_i=i\Delta t$.

Decrease the time interval $\Delta t$, and eventually you will approach the $45\text{ m}$ you are expecting.

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  • $\begingroup$ I like this perspective. $\endgroup$ – BMS Aug 21 '14 at 5:45

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