# AP Physics 1 Kinematics problem

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## Homework Statement

Luke Skywalker is making a diving run at the Death Star, approaching in his X-Wing at a rate of 500m/s. (The Death star is hollow, so it has almost no gravity.) Suddenly, at 800 meters from the Death Star, Luke fires his retro rockets. This creates a reverse acceleration of 160m/s^2, and when he reaches the nearest point, he fires his blasters. Then the X-wing begins moving away again, still powered by the retro rockets.
a.) How close will Luke be to the Death Star when he fires his blasters?

The ANswer is 19m, however, I keep getting it wrong. I am not sure where I made a mistake.

Do= 800m
D= ?
Vo= 500 m/s
V = 0
a = -160m/s^2

## Homework Equations

v2= 2(a)(d-do )+ vo2

3. The Attempt at a Solution
0= 2(-160)(d-800) + 5002
0= -320(d-800) + 250,000
0= -320d+256000 + 250000
320d= 506000
d=1581.25[/B]

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TSny
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0= 2(-160)(d-800) + 5002
How should d compare to 800? That is, is d larger or smaller than 800? With that in mind, what is the overall sign of the first term 2(-160)(d-800)? Can the equation be satisfied when the first term has this sign?

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