Deriving Rocket's position from acceleration.

[ShizukaSm]

Homework Statement

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:
<br /> \\<br /> a_x = 0.8\\<br /> a_y = -6.0 - 3.0t<br />
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)

Homework Equations

Just the equations already provided.

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:
<br /> \\<br /> x(t) = 0.4 t^2 + 600t + x_o\\<br /> y(t) = \frac{-1}{2}t^3 - 3t^2 + y_0<br />

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

 

[haruspex]

<br /> \\<br /> a_x = 0.8\\<br /> a_y = -6.0 - 3.0t<br />

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

 

[ShizukaSm]

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

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

 

[haruspex]

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

 

[ShizukaSm]

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

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.