# Determine time over known distance, constant acceleration and starting speed [closed]

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.

## closed as off-topic by Danu, user36790, Kyle Kanos, ACuriousMind♦, John DuffieldOct 8 '15 at 12:56

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• 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. – nluigi Oct 8 '15 at 8:39
• @nluigi can you please elaborate? I'm new to physics and don't know what that equation means – Peter-Pan Oct 8 '15 at 9:04
• 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 $\int dx = \int v\left(t\right) dt$ – nluigi Oct 8 '15 at 9:23

\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).