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This has been returned as homework and I have to make corrections. My teacher isn't very helpful and I can't afford a tutor.

When satellites circle closely around a planet or moon, the gravitational field surrounding the celestial body both increases the satellite’s velocity and changes its direction in an orbital move called a “slingshot.” Let’s say that a ship executing this maneuver has position $s(t) = t^3 - 2t^2 - 4t + 12$, where $t$ is in hours and $s(t)$ represents thousands of miles from Earth. What is the total distance traveled by the craft during the first five hours?

  • 1
    The position should be a 2-D or 3-D function. How come your position has only one dimension?2012-12-09
  • 0
    i don't know, is there no way to solve this, then, using applications of the fundamental theorem2012-12-09
  • 1
    there is.. but it will have nothing to do with slingshot trajectory2012-12-09
  • 1
    I have added the answer. Is this what you were looking for?2012-12-09
  • 0
    yes it helps, thanx2012-12-09

1 Answers 1

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The total distance traveled is the integral of speed i.e. $$D = \int_0^5|v(t)|dt,\mbox{ where } v(t) = s'(t)$$ $$v(t) = 3t^2-4t-4 = (3t+2)(t-2)$$ We need to find out $|v(t)|$. $v(t)$ has roots $-\frac{2}{3},2$. So, in the interval $[0,5]$, it changes sign at $t=2$. It is easy to see that $v(t)<0,t<2$ and $v(t)>0,t>2$

So, $$D = \int_0^2-v(t)dt + \int_2^5v(t)dt$$

It is easy to solve by hand. Wolfram gives the answer as 71 units, where units in this case are thousand miles.