# Intuitive explanation of the half in the $\frac{1}{2}at^2$ distance equation? [duplicate]

The full equation

$$Xf = X_o + V_o t + \frac{at^2}{2}$$

is integrated from the velocity function (which was integrated from constant acceleration function), right?

The problem is, I can't seem to put my mind around why the acceleration part needs to be halved.

Doesn't $at$ already measure the change in velocity over time? Why not just add that to the velocity then use the new V to measure Xf?

## marked as duplicate by Qmechanic♦Mar 7 at 18:49

• if you start from a standstill, the velocity is 0 at time=0 and V at time=t. In order to calculate how far it traveled. you have to calculate the average velocity over that time, which is v/2 which also is equal to at/2. So, for time =t, it travels at average velocity v/2=at/2 times t. – Peter R Feb 6 '16 at 6:17
• if u plot $v$ v/s $t$ in $v=u+gt$, it is a straight line, so the intercepts form a triangle, the formula you have written is the area of the triangle which is $\frac{1}{2}*base*height$ this is where $\frac{1}{2}$ comes from – Oswald Feb 6 '16 at 6:24
• Possible duplicates: physics.stackexchange.com/q/89590/2451 and links therein. – Qmechanic Jul 17 '16 at 19:32

Looking at the graph you can also see that the displacement is equal to the average velocity $\times$ time.

yes, the 1/2 is due to the double integral.

How $at$ evolves during the interval $[0,T]$ ? It's value is 0 at the begining, and $aT$ at the end. So the cumulated value in $v$ cannot be $aT^2$, or it would mean that the value was constantly equat to $aT$ during the interval.

When you sum-up varying quantities, you really have to compute the integral, not multiply the final value by $T$.

Taylor series of the space function $s(t)$:

$$s(t) = s_0 + s'(t_0)(t - t_0) + \frac{1}{2}s''(t_0)(t - t_0)^2$$

Now, having $s'(t_0) = v_0$ and $s'' = a$ you get:

$$s(t) = S_0 + v_0t + \frac{1}{2}at^2$$

assuming $t_0 = 0$

A simple intuitive explanation is that distance equals average speed times time. The average speed is $1/2(0+aT)$, and time is $T$ (assuming starting from rest, and constant acceleration).