Calculate displacement in position from knowing constant acceleration I have recently started studying physics at school, and my teacher went over the following equation without explaining about it too much:
$$s=\upsilon_{0}t+\frac{1}{2}a t^2
$$
I have wondered, why would this formula actually work? Is there an explanation for this?
 A: If you travel with constant speed $V$ for a time $T$, you will travel for $V\times T$ distance. Example: if your speed were $2$ m/s, and you were walking for $3$ seconds, you'd walk $2\times3 = 6$ meters.
Now, when calculating distance traveled while accelerating (or decelerating), we can approximate it if we split total time of travel into sub-intervals, and calculate sum of $V_i\times T_i$, where $V_i$ is speed at the beginning of $i$th sub-interval, and $T_i$ is its duration. 
The smaller intervals we take, the better is our approximation. And humans invented a way to calculate such sums using infinitely small sub-intervals -  definite integrals.
Imagine we have some function $f(x)$ and interval $[a, b]$. How we can calculate area of region between graph of $f(x)$ and $x$-axis on this interval? We can split interval $[a, b]$ in sub-intervals, and approximate the area with sum of areas of rectangles like shown on this image in Wikipedia. Sounds familiar?
If we have a formula $v(t)$ (speed from time), then we can calculate distance traveled during interval $[a, b]$ as area of the region bound by graph of $v$, $t$-axis, and two vertical lines at the ends of this interval.
If starting speed is $v_0$ and acceleration $a$ is constant, then speed in given moment of time $t$ is $v(t) = v_0 + a\times t$. If we start at time $0$, then at time $T$ distance traveled is
$$\int_0^T (v_0 + a\times t)dt = \left.(v_0\times t + a\times t^2/2)\right|_0^T = v_0\times T + a\times T^2/2$$ 
A: 
We know that the distance traveled is the area under the graph of the function $\:\upsilon(t)$. In this present question we do not need integrals, the areas are found by elementary geometry.    
