# Quick acceleration question

My physics is so rusty and I'm at work so I can't dig out my books but I have a quick question.

What is the formula to solve for time if one knows acceleration and distance?

Example: if one accelerates 10m/s/s for 1000km, how long does it take to cover the distance?

Anyone remember acceleration equations or have them handy?

Thanks

PS this isn't homework I'm trying to explain something to a co-worker and I'm just really rusty.

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$$d = Vi + \frac{1}{2}*a*t^2$$

However, in your case $$Vi = 0$$

cristo
Staff Emeritus
$$d = Vi + \frac{1}{2}*a*t^2$$

However, in your case $$Vi = 0$$
That's not the right general formula! If in the initial problem the car starts from rest it will not matter (since vi=0 in this case), but the correct formula is $$d=v_it+\frac{1}{2}at^2$$ Where d denotes the distance, vi the initial velocity, t the time, and a the acceleration. Note that you will need to convert the distance into metres.

Arg! I cant believe I forgot that, yeah it is Vi*t + .5at^2

My math is a little rusty too, but that would work out (since Vi is 0, Vi*t would equal 0) to be t= sqrt of 2d/a, yes??

cristo
Staff Emeritus
My math is a little rusty too, but that would work out (since Vi is 0, Vi*t would equal 0) to be t= sqrt of 2d/a, yes??
Yes, that's correct.

So then if one were to create a thruster that could continually supply the force needed for 10m/s/s of acceleration or just over 1g. in 2 years one would have traveled 2.11 light years and have exceeded the speed of light... right?

ly=9,460,073,432,260,800 meters
multiply that by 2.11 (half the distance to alpha centauri)
d=19,960,754,942.070.288
a=10

work that out and t=63,183,471s which is about 2 years.

obviously if one travels 2.11 light years in 2 years, one exceeded the speed of light (it that is possible)

but if it is and if one could create some nuclear drive or other perpetual thruster, one could accelerate at just over 1g halfway to alpha centauri, turn around and decelerate at just over -1g, making it to alpha centauri, which is 4.22light years away, in 4 years with "gravity" the entire time except during the rotation.

Right?

cristo
Staff Emeritus