# Catching up to object moving away

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1. May 6, 2015

### Dazed&Confused

1. The problem statement, all variables and given/known data
This is the final part of the problem so I won't give it exactly unless requested.

The relevant information thus far is that astronaut George is travelling at $(21i + 42j)ms^{-1}$ relative to astronaut Sandra and he is at position $(1260i + 2520j)m$ from her.

Sandra has a thruster pack capable of delivering a constant force of 120N in an arbitrary direction, but only has 30 s supply of fuel remaining. To help the still unconscious George she must reach him and ensure her velocity is equal to his as she get gets to him. She is prepared to use up all her fuel in order to do this as quickly as possible. In what directions and for how long should she apply the force from her thruster pack? If she starts at once, how long will it take her to reach George adopting this strategy?

Sandra has a mass of 60kg

2. Relevant equations
$r = \frac{1}{2}at^2 +v_0t + r_0$

3. The attempt at a solution

I'm going to work in one dimension, so replacing the position with 2817m and the velocity by $47ms^{-1}$. To find the initial time to apply the thruster pack, solve $$at -a(30 - t) = 46.7ms^{-1}$$ for t and you get 26.7 seconds. This is the given value.

The next part is where my value is completely different. To work out the time it takes her to travel to George, equate the total distance travelled by George with her and solve for the secondary $t'$ which is the difference between 26.7s and the total time.

Thus

$$\frac{1}{2}at^2 -\frac{1}{2}a(30-t)^2 + att' = 2817m + 47(t+t')$$

Solving for $t'$ gives the value 527s. The given answer for the total time of travel is 66.25s.

I have absolutely no idea how this could even be possible.

2. May 6, 2015

### SammyS

Staff Emeritus
Why subtract $\displaystyle\ \frac{1}{2}a(30-t)^2\$ ?

Isn't she moving forward as she decelerates?

3. May 6, 2015

### Dazed&Confused

I thought the differential equation for the last few seconds was $\frac{d^2x}{dt^2} = -2ms^{-2}$ and her velocity at that point in time was $at$.

4. May 6, 2015

### SammyS

Staff Emeritus
OK. -- I think.

Since your last equation and the parameters weren't explained well, I missed that.

So subtracting does seem to be correct.

5. May 6, 2015

### Dazed&Confused

Sorry about that. Also I've noticed I've used two values for the speed of George.

6. May 6, 2015

### SammyS

Staff Emeritus
Yup it's $\ 21\sqrt{5} \approx 46.96 \$ .

7. May 6, 2015

### Dazed&Confused

I realise I'm not being a 'good' thread starter. Should I add anything else?

Edit: they are actually both in a gravitational field $-g \hat{k}$, but this doesn't make any difference to their relative positions and velocity, right? ( because they are both accelerating at the same rate)

8. May 6, 2015

### SammyS

Staff Emeritus
Sandra accelerates (at a) for time, t . Then "coasts" for time, t', at constant speed at . Then decelerates for time, (30 - t) .

Plus whatever to match speed & distance.

9. May 6, 2015

### Dazed&Confused

Actually in my equation I included the deceleration in time $t'$ . You can think of it as coasting for time $t' -(30 - t)$ and then decelerating for $(30-t)$, but I grouped the terms together.