Moving a rocket between two points on a straight line, when to rotate from prograde to retrograde?

Question

Imagine you have a rocket and you want to move it from point a to point b. The flight plan is as follows:

1. Fire the rocket engine for a constant acceleration of 1 m/s^2 until x meters is covered (prograde burn)
2. Turn the rocket 180 degrees so that the rocket engine points retrograde. Note that while this happens the linear acceleration of the rocket is 0 but the velocity is still > 0, it still moves.
3. Fire the rocket engine for a constant deceleration of 1 m/s^2 until the last x meters is covered (retrograde burn)

It is logically that the burn in step 1 and 3 covers the same distance, x.

Say that the distance between a and b is 1000 meter and rotating the rocket takes 5 seconds. How do I approach finding a solution for x?

Normally I would start with computing the integral of the acceleration so I can find the distance the rocket has covered after several seconds. But in this case I don't know the distance (or time) after which the rocket should stop firing its rocket.

Please ignore other physical forces like gravity, drag, etc...

I've removed the image as it was indeed wrong and very confusing.

Answer 1

The trick is to figure out how long it will take to accelerate that distance and then use that to calculate the distance itself.

Let's lay out the variables:

• Uniform acceleration/deceleration $a$
• Known rotation time $t_R$
• Known total distance $d$
• Unknown acceleration/deceleration time $\tau$

The distance traversed during both acceleration and deceleration is $\frac{1}{2}a\tau^2$. Moreover, after acceleration, the ship is traveling at a speed of $a\tau$, so the distance traversed during rotation is $a\tau t_R$.

Putting all of this together yields the quadratic equation $$ d = a\tau^2 + a\tau t_R$$ which can be solved for $\tau$. Once $\tau$ is known, you can easily compute $\frac{1}{2}a\tau^2$.

Answer 2

Here is your picture, correctly labeled.

Under constant acceleration, the change in velocity is proportional to elapsed time, not proportional to distance traveled. If you actually plotted velocity over distance, the acceleration and deceleration segments would be curved. (Specifically, they would be arcs of parabolas with horizontal axes.)

So you actually spend some amount of time -- labeled $T_1$ on this relabeled diagram, since you did not specify a name for this period of time -- accelerating from zero velocity to the maximum forward velocity the rocket will reach. Then you spend $5$ seconds at that velocity, and then another $T_1$ seconds decelerating.

The distance traveled during any period of time is velocity integrated over time, which in this figure is represented by the area under the graph of velocity. So the total area under the velocity graph (the red and green lines) is $1000$. The distance traveled in the first $T_1$ seconds is $x$, which is the area of the triangle shown in the left side of the figure, and the area of the remaining region under the velocity graph is $1000 - x.$

Now that you know the correct place in the diagram to look for $x,$ can you solve the problem?