I'm trying to learn physics from the ground up because I didn't have it as a subject in high school. Currently I'm studying motion.

As far as I know and as far as my logic goes the formula for the distance traveled is as follows:


but in the current physics book I have it states that the formula for distance traveled is


Why is it so?

EDIT: It doesn't make sense to me... The second formula would state that $v = \frac{2s}{t}$

  • $\begingroup$ -1. What is the context of the equations? You cannot apply equations out of context. If you read the book it will tell you the context in which the equation applies. Neither of these equations is correct in all contexts. $\endgroup$ Nov 21, 2017 at 13:52
  • $\begingroup$ It was confusing for me, it defined the second formula for ""linear speeding up motion" The book is not in English. $\endgroup$
    – ChrisK
    Nov 21, 2017 at 13:59
  • $\begingroup$ "linear speeding up motion" is another way of saying "constant acceleration". $\endgroup$ Nov 21, 2017 at 14:06

2 Answers 2


The first equation assumes that the velocity $v$ is constant. In that case the distance travelled is simply the velocity multiplied by the time.

The second equation assumes that the acceleration is constant, so the velocity is increasing linearly with time:

$$ v = at $$

and it also assumes the object starts from rest i.e. $s = 0$ when $t = 0$.

There are several ways to understand the second equation, but the simplest is to say that if the object starts with velocity zero and after a time $t$ has velocity $v$, then the average velocity over this period was $\tfrac{1}{2}v$. The total distance travelled is the average velocity multiplied by time, so we end up with:

$$ s = \tfrac{1}{2}vt $$

A proper derivation of the second formula requires integral calculus, but there is an easy way to derive the formula graphically. If we graph velocity against time then the area under the graph is equal to the distance travelled. For constant velocity the graph looks like:

Constant velocity

It should be obvious that the area under the line (the shaded rectangle) is just $v$ times $t$ so we get:

$$ s = vt $$

If the velocity starts at zero and increases with time with constant acceleration then the graph looks like:

Constant acceleration

And now the area is the shaded triangle, which is half base times height or:

$$ s = \tfrac{1}{2}vt $$

  • 1
    $\begingroup$ Thank you very much! Visualizing it made it much simpler! $\endgroup$
    – ChrisK
    Nov 21, 2017 at 13:50

$s=vt$ holds when the velocity $v$ is constant. If it is not constant, one must start with $ds=v(t)dt$ and use $v$ as a function of time. Under the simplest of assumptions, the velocity $v(t)=at$, where $a$ is the acceleration assumed constant, so $$ ds=at dt\quad\Rightarrow\quad s=\frac{1}{2}at^2=\frac{1}{2}v(t)t\, . $$ The expression $ s=\frac{1}{2}at^2$ is much more prevalent precisely because of the possible confusion between $s=vt$ for $v$ constant and $s=\frac{1}{2}v(t)t$ for $v(t)=at$ increasing with $t$. Note that the full expression for $s$, assuming constant acceleration $a$ is $$ s=s_0+v_0t+\frac{1}{2}at^2 $$ with $s_0$ and $v_0$ the initial position and velocities, respectively.


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