How do you find a constant acceleration if distance and time are known?

Question

I Know Distance.
I Know Time.

I want to know what the constant acceleration and decelleration would be if I left my destination and arrived at my destination at 0m/s.

For instance.
au = Astronamical Unit
ly= Light Year ( 63240au)

Distance = 10ly (632400au)
Time = 180 seconds

What is your acceleration/deceleration in AU/S/S

What equation would you need to work out acceleration if you accelerated and then decelerated at a constant rate for half of the journey?

Ok, thanks for the warnings on the speed.
We're obviously using a none classical method of transportation.

However back to the math.

How do I re-arrange the equation S = UT + 1/2 AT^2
When T is not known, but when S, U and A are?

Ok, thanks for the warnings on the speed.
We're obviously using a none classical method of transportation.

However back to the math.

How do I re-arrange the equation S = UT + 1/2 AT^2
When T is not known, but when S, U and A are?

Answer 1

s = s0 + v0t + (1/2)at^2
s0 = 0
v0 = 0
a = 2s/t^2
Since you are doing a flip-over, halve the original distance. Using the numbers given,
a = 632400/180^2 = 19.51852 AU/s^2

This is a really hellish acceleration. I'm not sure there is any material that can withstand it.

Answer 2

Using classical mechanics VUAST

You have
U = 0
V = ?
A= ?
S = 5ly (flip-over point)
T = 90 seconds
You need A so you need the equation that does not have V in it

S = UT + 1/2 AT^2
And putting in the figures A = 10/8100 Ly/sec^2

Please note that if you work out V you get V = 10/90 Ly/sec which is a lot quicker than the speed of light and so impossible!

You must therefore start using relatavistic mechanics - which I've forgotton as its so long ago.