# Stopping Distance (frictionless)

Assuming I have a body travelling in space at a rate of $1000~\text{m/s}$. Let's also assume my maximum deceleration speed is $10~\text{m/s}^2$. How can I calculate the minimum stopping distance of the body?

All the formulas I can find seem to require either time or distance, but not one or the other.

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If the speed is $1000 m/s$ and the deceleration is $10 m/s^2$, it will take $100 s$ to stop. The average speed in that time is $500 m/s$, so the distance traveled is

$$500m/s*100s = 5*10^4m$$

Working through the same logic with an initial speed $v$ and a deceleration $a$, the final distance $d$ traveled before stopping is

$$d = v_{avg}*t = (v/2)*(v/a) = \frac{v^2}{2a}$$

This formula becomes more interesting when you learn a bit more physics because it's simple example of the work-energy theorem.

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Thanks for the actual formula. – Timothy Baldridge Jan 25 '11 at 4:46

The formula you want is

$$v_f^2 = v_i^2 + 2a(x_f - x_i)$$

It's one of the basic kinematic formulas taught in high school (or even middle school) physics classes.

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@David Well, I think that's a new record. You beat me by three seconds. – Mark Eichenlaub Jan 25 '11 at 4:44
Thanks, it's been a few years since I've done this math. Don't use something for almost a decade and you start to get rusty. – Timothy Baldridge Jan 25 '11 at 4:45
@Mark: well I haven't really been paying attention to the timing, but yeah, that is pretty close. Now you've got me wondering about the distribution of time differences between consecutive answers posted to the same question ;-) – David Z Jan 25 '11 at 4:51
@David: that's interesting, I'd love to see data on it! Is it possible to get high resolution time data for answer posting? We could plot the intervals on a histogram and make a meta thread. – Colin K Jan 25 '11 at 6:34
@Colin: The Stack Exchange API provides posting times down to the second. You probably couldn't get better resolution than that (since I suspect it isn't stored in the database to any higher precision), but I think that should be enough since most of these time intervals are measured in hundreds or thousands of seconds. – David Z Jan 25 '11 at 6:41

Another equation of motion problem,very easy.

$$v^2 = u^2 + 2.a.s$$ where $$v = \textbf{final velocity}$$ , $$u = \textbf{initial velocity}$$ , $$a=\textbf{ acceleration or in this case negative acceleration}$$ , $$s = displacement$$ . This equation is time independent. Now, to find $s$ , put the values: $$s = \frac{-1000^2}{2.(-10)} \implies s = 50000~\text{m}$$ . Simple,right?

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Ok, this is similar to @DavidZ but rarely do such easy questions come to be answered by a novice like me! – MAFIA36790 Oct 8 '14 at 6:56