Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have recently started studying physics at school, and my teacher went over the following equation without explaining about it too much:

$$d~=~vt+\frac{1}{2}a t^2.$$

I have wondered, why would this formula actually work? Is there an explanation for this?

share|cite|improve this question
Note that this site supports MathJax for equation rendering. I've edited your post to use it. Look at the FAQ if you want to see how it works. – Michael Brown Mar 15 '13 at 12:12
Do you have any knowledge of differential equations? Because that would be required to explain mathematically what the formula comes from – Michiel Mar 15 '13 at 12:12
@michielm I do have knowledge in differential equations, go ahead :) – arielschon12 Mar 15 '13 at 17:17
up vote 3 down vote accepted

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

$$\int_0^T (v_0 + a\times t)dt = \left.(v_0\times t + a\times t^2/2)\right|_0^T = v_0\times T + a\times T^2/2$$

share|cite|improve this answer
Thanks a lot, this answer is perfect! – arielschon12 Mar 15 '13 at 17:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.