Problem: I have an object covering a specific distance (say 100m) at a starting speed (say 10m/s) and a constant acceleration of 1m/s/s.

Which equation can I use determine how long it will take the object to reach the 100 meter mark?

I've seen the equation $x = \frac{1}{2}at^2$ used for a similar problem but with starting speed of 0.


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  • 2
    $\begingroup$ The clue is in constant acceleration: $$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$$ Integrate, apply relevant initial conditions and you are good to go. $\endgroup$ – nluigi Oct 8 '15 at 8:39
  • $\begingroup$ @nluigi can you please elaborate? I'm new to physics and don't know what that equation means $\endgroup$ – Peter-Pan Oct 8 '15 at 9:04
  • $\begingroup$ Acceleration a is the rate of change of velocity v with time t, i.e. $a=\frac{dv}{dt}$. Velocity is the rate of change of position x with time, i.e. $v=\frac{dx}{dt}$. In the same way, acceleration is the rate of change of the rate of change of position, i.e. $a=\frac{d}{dt}\left(\frac{dx}{dt}\right)=\frac{d^2x}{dt^2}$. Now how familiar are you with calculus? What is the integral of a constant, e.g. $\int dv = \int a dt$? Determine this integral, find the integration constant by applying the initial speed condition. Then go onto position x by using $\int dx = \int v\left(t\right) dt$ $\endgroup$ – nluigi Oct 8 '15 at 9:23

At school physics level the vast majority of systems you need to analyse are described by one of the SUVAT equations. Sadly the Wikipedia article on these got deleted, but Googling SUVAT equations will find you lots of related articles. Searching this site will find lots of related questions.

The equations are all derived from Newton's second law as discussed by nluigi in the comments. There are three equations, though they are all different versions of the same thing:

$$\begin{align} v &= u + at \\ s &= ut + \tfrac{1}{2}at^2 \\ v^2 &= u^2 + 2as \end{align}$$

where $s$ is distance, $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration and $t$ is time (hence the acronym SUVAT).


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