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This question already has an answer here:

My question is, we can calculate distance of free falling objects with the equation of $d = \frac{1}{2}gt^2$ where $d$ is distance in meter, $t$ is time in seconds, but I wonder how we achieved this equation. I know that the object's velocity isn't constant, the object is accelerating. So we can't calculate it via $x = vt$. So, can anyone explain me this equation? Thanks in advance! :)

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marked as duplicate by ACuriousMind, Bill N, Sebastian Riese, Qmechanic Sep 18 '15 at 15:39

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  • $\begingroup$ Find any college physics text used in a calculus based physics course. The derivation is in there. Knight, Serway, Halliday, Young ... all these authors have books that can help you. $\endgroup$ – Bill N Sep 18 '15 at 12:45
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/89590/2451 and links therein. $\endgroup$ – Qmechanic Sep 18 '15 at 15:38
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The answer by ryanp16 gives a great derivation of the equation you're asking about using calculus, and in fact, that is the approach that I would have taken had I not seen his answer. However, if you're not familiar with calculus, there's a second, algebraic approach that you can use to arrive at the same conclusion. I like to take this approach with high school students who have not learned calculus yet.

First, recall that the average speed of an object is given by $$\bar{v}=\frac{\Delta x}{\Delta t}$$

For an object travelling at constant speed, this average speed is equal to the instantaneous speed of the object at any point during the interval. However, if the object is changing velocity at a constant rate (constant acceleration), the average velocity can also be described by $$\bar{v}=\frac{v_0+v_f}{2}$$

Just for kicks and giggles, I will set these two equations equal to each other (because their LHSs both equal $\bar{v}$). Then, for reasons unknown yet, I will solve for $v_f$. $$\frac{\Delta x}{\Delta t}=\frac{v_0+v_f}{2}$$ and then, when solved for $v_f$ $$\frac{2\Delta x}{\Delta t}-v_0=v_f$$

Don't worry, we'll see the pieces come together soon. At first glance, it may not be obvious why this relationship is helpful to us, but remember, we have another equation that relates final and initial velocities, as well as acceleration. Recall the kinematic equation $$v_f=v_0+at$$ which is basically a rearranged version of the equation definition of acceleration. Now, that kinematic equation, as well as the equation I derived above, can be set equal to each other, because both $\frac{2\Delta x}{\Delta t}-v_0$ and $v_0+at$ are equal to $v_f$. So we get $$\frac{2\Delta x}{\Delta t}-v_0=v_0+at$$ This is pretty cool, because now when I solve for $\Delta x$ I get $$\Delta x= v_0t+\frac{1}{2}at^2$$ You chose to use $d$ as your symbol for displacement, which is fine. Additionally, if your initial velocity is zero, and your acceleration, $a$ is the acceleration due to gravity, $g$ you'll get your equation $$d=\frac{1}{2}gt^2$$

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  • $\begingroup$ That's definitely useful for the calculus-impaired. Nicely done! $\endgroup$ – Gert Sep 18 '15 at 13:11
  • $\begingroup$ That's a good second option, thank you for clearing :) $\endgroup$ – ysn_akst Sep 18 '15 at 14:34
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Acceleration $a$ is defined as the rate of change of velocity $v$ with respect to time $t$, or $a=\frac{dv}{dt}$. For constant acceleration we can integrate both sides with respect to $t$ to obtain $v=u+at$ where $u$ is the velocity at our initial time $t=0$.

Since velocity $v$ is defined as the rate of change of displacement $s$ with respect to time we have $\frac{ds}{dt}=v=u+at$. Integrating both sides with respect to $t$ we have $s=ut+\frac{1}{2}at^2$, with the initial condition $s=0$ at $t=0$.

In your example the object is dropped from rest so $u=0$ and $a=g$. So the distance traveled in $t$ seconds is given by $s=\frac{1}{2}at^2$.

By the way, this assumes there are are no other forces such as wind resistance etc acting on the falling body, which would be the case in real life. Then the acceleration isn't constant so the integration is slightly less trivial.

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  • $\begingroup$ Your answer is really good, thank you. It was haunting my mind $\endgroup$ – ysn_akst Sep 18 '15 at 11:25

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