Solving 2007 F=ma Physics Olympiad Problems: AP Phys. C Study Guide

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The discussion focuses on solving problems from the 2007 Physics Olympiad F=ma test, with the user sharing their attempts and seeking feedback. They correctly identify when an object crosses the x-axis and calculate average velocity using derivatives, arriving at the right answers for the first two problems. For the third problem, they accurately determine the distance fallen during the first and second seconds, confirming their reasoning aligns with the physics principles involved. However, there is a correction regarding the interpretation of the equations, particularly in identifying which components relate to the x and y axes. Overall, the user is actively engaging with the material and refining their understanding of physics concepts.
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I'm going through the 2007 Physics Olympiad F=ma test tonight (first 27 questions). It's available http://www.aapt.org/Contests/upload/olympiad_2007_fnet_ma.pdf" , if you're wondering. I'll post my attempts here as I get to them. Tell me what I'm doing right or wrong. I don't really know this stuff, I'm taking AP Phys. C Mechanics right now.


1. An object moves in two dimensions according to \vec{r}(t) = (4.0t^{2}-9.0)î + (2.0t-5.0)ĵ. where r is in meters and t in seconds. When does the object cross the x-axis?
(a) 0.0 s (b) 0.4 s (c) 0.6 s (d) 1.5 s (e) 2.5 s
From what I have picked up, i and j represent x and y in the coordinate plane (and k is z). I would think that it would cross the x-axis when the stuff next to the y becomes zero. 4.0t^{2}-9=0 when t=1.5. So, my answer is d.

3. The coordinate of an object is given as a function of time by x = 8t - 3t^{2}, where x is in meters and t is in seconds. Its average velocity over the interval from t = 1 to t = 2s is
(a) -2 m/s (b) -1 m/s (c) -0.5 m/s (d) 0.5 m/s (e) 1 m/s
Well, velocity is just the derivative of position, so it would be given by v=8-6t. Since that's linear, the average of that is at the point t=1.5s, which means the answer is -1 m/s, b.

4. An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next second of time?
(a) h (b) 2h (c) 3h (d) 4h (e) h^2
Acceleration is 9.8 m/s^s down, and position is the double integral of that which means it's quadratic. So, in the first second it should have fallen 1^{2} h. At the end of the second second, it should be at 2^{2} h. 4-1=3, the answer is c.
 
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That looks ok.

I would treat the #3 differently. V = Δx/Δt

(4 - 5)/ 1 = -1
 
1. You are right. But the "stuff' next to y is 2t-5 and not 4t^2 - 9. You found out when it'll cross the y-axis.

4. This particular answer is right. But i don't think this analogy would work if the body was not thrown from rest. In that case the answer would depend on the initial velocity.
 
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