# Intuition behind the $\frac{1}{2}$-fraction in $s(t) = \frac{1}{2}v(t)t$ for motions with constant acceleration

I'm refreshing my mechanics knowledge and have a question.

Say we have an object moving with a constant acceleration $a$ moving in one dimension. Furthermore, $v_0 =0$, $s_0 = 0$ and $t_0 = 0$ (with $s$ being the traveled distance).

Then the speed $v$ of the object at time $t$ is given by $$v(t) = at$$

The traveled distance $s$ at time $t$ is given by $$s(t) = \int v(t)dt = \frac{1}{2}at^2$$ This can be written as $$s(t)=\frac{1}{2}at^2 = \frac{1}{2}(at)t = \frac{1}{2}v(t)t$$

So far so good. But what I forgot is, how do you interpreted the $\frac{1}{2}$-fraction intuitively? In other words, why is the current traveled distance given by half the amount of the current speed times the time?

• Yes, I used the notation $v(t)$ to indicate that $v$ is dependent on $t$. And you're right, the integration constant is zero. :) – Kevin Sep 7 '16 at 14:28
That expression for $v(t)$ is the current velocity at time $t$, which increases linearly in time (starting at 0 meters per second at time 0 seconds). The displacement can also be calculated with the average velocity times the total time. But the average velocity from zero to $t$ is half the current velocity.
Imagine the integral s(t) geometrically. The area you calculate is a right triangle, with sides of length t and v(t). The area of this triangle is therefore $\frac{1}{2} v(t) t$.