# Constant Acceleration in Space. How much time for a given distance?

I have been researching a lot and can't make any sense of it all, so I am trying my luck here.

I want to program a simple formula that gives me the travel time in space.

I have a constant thrust of let's call it 'a'. The spaceship starts to decelerate at the midpoint, 'm'.

the total distance is 'd', so m=d/2

How long does the complete voyage take and what is the speed at the midpoint and at any given time, please?

No factoring of lightspeed/relativity stuff, please. ;)

Help would be much appreciated. Thanks.

• do you want constant acceleration, or constant thrust? These give very different results. Jan 17 '19 at 17:01
• Jan 17 '19 at 17:32
• well, first thank you all for replying. Ok, to clarify, I have constant mass in the ship and an engine that produces a constant thrust. Is the acceleration instant or would it build up? What would be the graph like for such a thing? Jan 18 '19 at 14:38
• @SirSlarti The acceleration will be low at first, and then increase as your ship weighs less due to consuming propellant. Mar 15 '20 at 1:01
• For really high speeds you need to use the relativistic formulas: en.wikipedia.org/wiki/Space_travel_using_constant_acceleration x( t)= {c^2/a} ( sqrt(1+(at/c)^2 ) -1 ); v(t)= {at} / { sqrt(1+(at/c)^2) };
– HJC
Jan 17 at 16:40

for a straight line, the speed at the midpoint is $$a\frac{t}{2}$$. The average speed is half that $$v_{avg}=\frac{a\frac{t}{2}}2 =a\frac{t}{4}$$ and the total time is $$t=\frac{d}{v_{avg}}$$ and so \begin{align} t=&\frac{d}{a\frac{t}{4}}\\ t=&\frac{4.d}{a.t}\\ t^2=&\frac{4d}{a}\\ t=&2\sqrt{\frac{d}{a}} \end{align}
The speed at an instant in time is $$v_i=a\left(\frac{t}2-ABS{\left(\frac{t}2-i\right)}\right)$$
• no, @SirSlarti, that takes it into account -- as should be clear from the maths: average speed with constant acceleration throughout the trip would be $v=\frac{a.t}2$