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hsbeth

Quick emergency Physics question?

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Dd is struggling with a physics problem that is very similar to this one linked:http://www.chegg.com/homework-help/questions-and-answers/a-rocket-powered-hockey-puck-moves-on-a-horizontal-frictionlesstable-the-figure-shows-grap-q452766

 

Dd's problem is identical except that the slope for the vx graph is positive. Her problem reads:

 

A rocket powered hockey puck moves on a horizontal frictionless table. The puck starts at the origin. What is the magnitude of the puck's acceleration at t=5 s

 

She has determined that ax is 2m/s^2 and that ay is 1m/s^2

 

 

Her  prof. is two lectures behind as he's been interviewing for another positon :/ and apparently has not been keeping his office hours.

 

Grateful for any help!

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Ds says to search the magnitude of vectors. He says that magnitude equals the length of the vector and that by the Pythagorean theorem you can find the magnitude equals (ax^2+ay^2)^.5=(2^2+1^2)^.5=(5)^.5

 

Eta:he was walking out the door and this was a 30 sec answer.....I hope it helps.

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Looks like she's found the correct values of the components ax and ay of the acceleration vector at t=5 s.

Since acceleration is a vector quantity, its magnitude at that instant is then equal to sqrt(ax2 + ay2), where "sqrt" stands for square root. (The magnitude of any vector quantity is the length of that vector. One way to find this length is to use the Pythagorean theorem. The vector is the hypotenuse, and its x and y-components are the two legs of the right triangle.)

 

ETA: cross-posted with 8Fill. I agree with her son that the answer is the square root of 5.

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the components ax and ay are the short sides of the right triangle that has the magnitude a (ie. length) of the vector as hypotenuse

use Pythagoras.

 

typed out long winded explanation of finding the components.. duh. She did that correctly

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Thank you! Thank you!!! Her father suggested a similar approach and she insisted he was wrong. I think she was thrown b/c of the specific time? She finally jumped in the shower after working on this for some 90 minutes. Will run all this by her as soon as she gets out.

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Thank you! Thank you!!! Her father suggested a similar approach and she insisted he was wrong. I think she was thrown b/c of the specific time?

 

the time IS unnecessary information! Since both vx (t) and vy(t) are straight lines, the x- and y-components of the acceleration are both constant throughout the motion.

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Thank you! Thank you!!! Her father suggested a similar approach and she insisted he was wrong. I think she was thrown b/c of the specific time? She finally jumped in the shower after working on this for some 90 minutes. Will run all this by her as soon as she gets out.

Tell her that when I asked my ds he said that the question was meant to annoy bc acceleration is constant and t=5 is irrelevant. (And I don't have a clue about any of it.)

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Tell her that when I asked my ds he said that the question was meant to annoy bc acceleration is constant and t=5 is irrelevant. (And I don't have a clue about any of it.)

 

If you look at the graphs, you'll have a clue, too :-)

Acceleration is simply velocity change per time. Since velocity vs time is a straight line, every second the velocity changes by exactly the same amount - so the acceleration is the same wherever you look. Geometrically, acceleration can be seen as rise over run, i.e. slope, of the v-t graphs.

 

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If you look at the graphs, you'll have a clue, too :-)

Acceleration is simply velocity change per time. Since velocity vs time is a straight line, every second the velocity changes by exactly the same amount - so the acceleration is the same wherever you look. Geometrically, acceleration can be seen as rise over run, i.e. slope, of the v-t graphs.

 

I honestly didn't even look at the problem. ;) it was just easier to ask him. :) but, thank you for the explanation.

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