# Deriving Rocket's position from acceleration.

1. Dec 9, 2012

### ShizukaSm

1. The problem statement, all variables and given/known data
There's a short introduction saying that honing missiles can determine their position by utilizing their acceleration, then the problem says: Suppose that the missile's acceleration obeys the following equations:
$$\\ a_x = 0.8\\ a_y = -6.0 - 3.0t$$
Knowing that, in $t = 0$, $v_x = 600 km/h$ and $v_y = 0$, calculate the missile's displacement in ten seconds. (Answer is 2.2km)

2. Relevant equations

Just the equations already provided.

3. The attempt at a solution
Alright, so, what I tried to do was integrating the acceleration two consecutive times to arrive at an equation for x(t) and y(t), which gave me:
$$\\ x(t) = 0.4 t^2 + 600t + x_o\\ y(t) = \frac{-1}{2}t^3 - 3t^2 + y_0$$

Then I attempted to get the displacement of both axes(Using time = 1/360h, since I have to convert 10 seconds to hours):
$$\\ x(10)-x(0) = 1.67km\\ y(10)-y(0) = -2.3*10^{-5} km$$
And then, by calculating$sqrt(x^2 + y^2)$I obviously get pretty much 1.67km, which is wrong.

2. Dec 9, 2012

### haruspex

What are the units of distance and time in those equations?

3. Dec 10, 2012

### ShizukaSm

That's a great question. I just realized it says in parenthesis (SI Units), so m/s. In other words, I shouldn't convert time to seconds, I should convert km/h to m/s. However, by converting 600 km/h to 166.67 m/s and doing the necessary calculations I still obtain only 1884.26 m(With x =1706 and y = -800) : S

4. Dec 10, 2012

### haruspex

ax = 0.8 looks suspicious. Why so much less than ay? If you make it 8.0 you get the book answer.

5. Dec 10, 2012

### ShizukaSm

Wow, you're absolutely right! Wonderful! Thanks.

It definitely says 0.8 and It's actually an online list of exercises, not a textbook, so I find very possible that the person that typed it made a mistake.