# How is this equation derived?

1. Jul 10, 2004

### Esran

$$d=v_0t+\frac{1}{2}at^2$$

Why is it $$\frac{1}{2}at^2$$? Why not $$at^2$$ instead? I know that the reason is probably pretty obvious, but I'm having trouble finding it.

Thanks for your help.

Last edited by a moderator: Jul 10, 2004
2. Jul 10, 2004

### robphy

Integrate $$v=v_0 + a t$$ with respect to $$t$$.

3. Jul 10, 2004

### selfAdjoint

Staff Emeritus
In case you haven't had integration yet, try this.

Since a is constant, you can graph v in the t,v plane as a straight line slanting up from t=0, v=v0 to some generic point t=t, v=at. Drop a vertical from there and run a horizontal line from the initial point; you have a right triangle and the area of that triangle is the distance made good. But what is the area of a triangle? It is the base (elapsed time=t) times 1/2 the altitude (change in v). But the change in v is at - v0, therefore the distance is given by your formula. Draw the picture.

Last edited: Jul 10, 2004
4. Jul 11, 2004

### ArmoSkater87

I think its a simple substitution using another constant acceleration equation.

$$v=v_0 + a t$$

$$V^2 = V_0^2 + 2ad$$

$$V=\sqrt{V_0^2 + 2ad}$$ sub. into the first equation

$$\sqrt{V_0^2 + 2ad} = V_0 + at$$ Square both sides

$$V_0^2 + 2ad = V_0^2 + 2V_0at + a^2t^2$$ simplify and divide both sides by 2a

$$d=v_0t+\frac{1}{2}at^2$$

Last edited: Jul 11, 2004
5. Jul 11, 2004

### gerben

a is the acceleration, which is the change of the velocity per second. You started with velocity $v_{start} = v_0$, so at the end after $t$ seconds the velocity is:
$$v_{end} = v_0 + at$$

The average velocity during these $t$ seconds was:

$$\frac{v_{start} + v_{end}}{2} = \frac{v_0 + v_0 + at}{2} = \frac{2v_0 + at}{2} = v_0 + \frac{1}{2}at$$

To get the distance travelled during this time you have to multiply this by the time:
$$d=(v_0 + \frac{1}{2}at)t = v_0t + \frac{1}{2}at^2$$

6. Jul 16, 2004

### Nenad

I tried proving this. If you integrate, it is easy. But I tried it the old fashioned way and Im getting the same problem as Esran.
What I did was I had 2 equations to work with. a = (v - vo) / t, and v = d/t

a = ((d/t) - vo)/t
at = (d/t) - vo
at + vo = dt
at^2 + vot = d
where does the 1/2 come in?

7. Jul 16, 2004

### ZapperZ

Staff Emeritus
The error comes in because you let v=d/t, which is not the instantaneous velocity, but rather the AVERAGE velocity since v isn't a constant (there is an "a", remember?).

Zz.

8. Jul 17, 2004

### Nenad

ohh, k , thanx

9. Jul 17, 2004

### BobG

Self Adjoint explained the math to it.

The reason behind it is this. Acceleration is fow fast your velocity is increasing. If you started out from rest and accelerated at 10 m/s^2, you would be travelling 10 m/s after one second. But you didn't travel 10 m/s for the entire second. You were travelling 0 m/s at time 0, 1 m/s after .1 seconds, 2 m/s after .2 seconds, etc. Your average velocity was 5 m/s (1/2 your acceleration).

This continues on even after the first second, except now you're starting at 10 m/s and reach 20 m/s by the end of the 2nd second. In other words, your average velocity for the 2nd second is 15 m/s. You total distance, so far, is 20 meters (5 meters the first second and 15 meters the 2nd second). You can keep going on second by second like this. The 1/2 a accounts for the non-constant velocity while the t^2 accounts for the accumulating distance from second to second.

10. Jul 17, 2004

### Entropy

Yeah back when I was a freshmen and just started physics I wondered what was up with that equation. Since I learned a little calculus I understand.

11. Jul 18, 2004

### JohnDubYa

I vote BobG has having the best answer.

12. Jul 19, 2004

### ArmoSkater87

its good, but i like gerben's better

13. Jul 19, 2004

### robphy

While this proof works, it should be noted that the "average velocity during these $t$ seconds was $\frac{v_{start} + v_{end}}{2}$" is a consequence of the acceleration being constant. Recall that "average velocity" means "time-weighted average of the velocities", not merely a straight average of the velocities at the endpoints. Indeed, following the definitions
\begin{align*} v_{avg} &\equiv\frac{x_f-x_0}{t_f-t_0}\\ &=\frac{\left[x_0+v_0t+\frac{1}{2}at^2\right]-x_0}{\left[ t \right]}\\&=v_0+\frac{1}{2}at \\ &= v_0+\frac{1}{2}\left( v_f-v_0 \right) = \frac{1}{2}\left(v_f+v_0 \right) \end{align *}

In some texts, the above expression for average velocity as one-half the sum of initial and final velocities is justified by drawing a velocity-vs-time graph and essentially arguing that the area under the red curve is equal to the area under the green curve. Of course, the area under the red curve is the displacement gained over the interval: the lower rectangle has area $$v_0 t$$ and the triangle (with height $v_f-v_0=at$) has area $\frac{1}{2} (at) t$:

$$\begin{picture}(100,100)(0,0) \put(0,0){\vector(1,0){100}} \put(0,0){\vector(0,1){100}} \put(0,30){\textcolor{red}{\line(3,1){100}}} \put(0,30){\textcolor{red}{v{\scriptsize \textcolor{red}{0} }}} \put(0,47){\textcolor{green}{\line(1,0){100}}} \put(0,50){\textcolor{green}v{\scriptsize \textcolor{green}{avg}}} \put(0,30){\line(1,0){100}} \put(100,64){\line(0,-1){64}} \put(0,64){\textcolor{red}{v{\scriptsize \textcolor{red}{f} }}} \put(95,0){t} \end{picture}$$

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