# DE: Models of Motion

1. Sep 29, 2008

### rocomath

A rocket is fired vertically and ascends with constant acceleration a=100m/s^2 for 1min. At that point, the rocket motor shuts off and the rocket continues upward under the influence of gravity.

a) Find the maximum altitude acquired by the rocket

b) The total time elapsed from the take-off until the rocket returns to the earth

Ignore air resistance.

a) The maximum height can be found through setting velocity equal to zero, and the integral of acceleration is velocity.

$$\int_0^Ta(t)dt=\int_0^{60}a(t)dt+\int_{60}^Ta(t)dt=0$$

$$\int_0^Ta(t)dt=\int_0^{60}100dt-\int_{60}^T9.8dt=0$$

T=672.244898s

Is (a) good so far?

Last edited: Sep 29, 2008
2. Sep 29, 2008

### gabbagabbahey

(a) looks good so far.

3. Sep 29, 2008

### rocomath

yay! Thanks for the confirmation.

4. Oct 6, 2008

### rocomath

Ok, so I have the time when my rocket is at it's highest. I know the position function equation, but I can't use it since my acceleration isn't constant.

Can I modify it such that $$x_1(t)$$ has g=100, and $$x_2(t)$$ has g=9.8, and the maximum height is just $$x_1+x_2$$???

5. Oct 6, 2008

### gabbagabbahey

Well $a(t)=\dot{v}(t)$ so why not integrate a(t') from t'=0 to t'=t to find the speed of the particle at a time t the same way you did to find v(T). Treat two cases, t<60min and t>60. this should give you the piecewise function for v(t). Then, repeat the process to find x(t) and finally plug in to the equation to find x(T).

Last edited: Oct 6, 2008
6. Oct 6, 2008

### rocomath

Oh ok, so just keep on integrating! Let me try it.

Thanks :)

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