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So I have a body which moves straight on a line. The body accelerates from 0 m/s to some velocity V, the acceleration is constant and I know that after T seconds it reaches its final velocity V. Is there a way to find the distance it travelled during those T seconds?

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Yes you can. There are five equations that you can quote (derivation below) for motion in a straight line under constant acceleration that relate the quantities: displacement, $s$, initial velocity, $u$, final velocity, $v$, acceleration, $a$, and time, $t$. They often referred to as SUVAT equations for obvious reasons!

The equation that applies to your problem is:

$$s = \frac{1}{2}(u + v)t. \tag{1}\label{1}$$


To derive this, consider a graph of speed against time for your situation:

graph source: https://www.ncetm.org.uk/resources/52734

Displacement is the integral of velocity, which is the area under this graph. Clearly this area can be expressed as $A = ut + \frac{1}{2}(v - u)t = \frac{1}{2}(u+v)t$, which is Equation \eqref{1}.

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  • $\begingroup$ Thanks, this is what I was looking for! $\endgroup$ – Xriuk May 7 at 13:45
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    $\begingroup$ @Xriuk No problem! Note that this formula is for displacement though, not distance, so if the velocity were to transition from positive to negative (or negative to positive), you would not have the "total distance travelled", but the displacement. In this case you would have to split the problem up into two sections. $\endgroup$ – Joe Iddon May 7 at 13:53
  • $\begingroup$ sorry, I didn't quite get the difference between displacement and distance? If both v and u are positive (or both negative?) the displacement equals the distance? $\endgroup$ – Xriuk May 7 at 14:41
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    $\begingroup$ Yes, correct. However if the initial is positive and the final is negative, then this calculation will give you your "final position" after moving with those velocities. I.e. although you may have reached a distance of, say, $5m$ from the origin, if you then turned around and came back $2m$, this calculation would report that your displacement is $3m$, whereas the total distance travelled would be $7m$. $\endgroup$ – Joe Iddon May 7 at 19:06
  • $\begingroup$ Oh, I get it now, thanks! $\endgroup$ – Xriuk May 8 at 8:44

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