View Full Version : The impossibe problem
Mike Hunt
Sep7-08, 11:48 AM
Okay so its not really impossible but I don't understand it.
1. The problem statement, all variables and given/known data
http://i528.photobucket.com/albums/dd323/shazamanam/Untitled.jpg
2. Relevant equations
Velocity is change in position over change in time.
Acceleration is change in velocity over change in time.
3. The attempt at a solution
My attempts are in the picture. I only got one box right. Please give me a hint or two please?
LowlyPion
Sep7-08, 12:00 PM
Okay so its not really impossible but I don't understand it.
Velocity is change in position over change in time.
Acceleration is change in velocity over change in time.
3. The attempt at a solution
My attempts are in the picture. I only got one box right. Please give me a hint or two please?
Welcome to PF.
Go box by box then. You seem to have mostly the right idea with these, maybe just a few subtleties away from getting full credit.
So explain box 1 ax < 0.
Edit: A more efficient approach, if you prefer, might be to describe what you think is happening at each of the 7 points on the curve, and then the answers for each box will be simply a matter of recording them correctly?
loonychune
Sep7-08, 12:06 PM
Velocity is change in position over change in time.
Acceleration is change in velocity over change in time.
Then acceleration is the rate of change of the rate of change of position with time.
You're essentially looking to see whether the change of position with time, in-around some time, is positive or negative..
Ygggdrasil
Sep7-08, 12:21 PM
Velocity is the derivative of the position graph. Drawing this out and then drawing the acceleration graph (derivative of the velocity v. time graph) may be helpful.
Mike Hunt
Sep7-08, 01:06 PM
http://i528.photobucket.com/albums/dd323/shazamanam/sdtrhyge.jpg
For the first 3 boxes these are my answers that I think are right. I get 5 chances to submit the right answer and I have submitted the first 3 four times. This is my last chance. Ignore the red Xes because I have not submitted this answer yet. Tell me does this look correct to you?
jmartella
Sep7-08, 01:34 PM
http://i528.photobucket.com/albums/dd323/shazamanam/sdtrhyge.jpg
For the first 3 boxes these are my answers that I think are right. I get 5 chances to submit the right answer and I have submitted the first 3 four times. This is my last chance. Ignore the red Xes because I have not submitted this answer yet. Tell me does this look correct to you?
EDIT: Sorry, I thought the graph was a velocity vs. time.
Redbelly98
Sep7-08, 01:40 PM
Acceleration is the 2nd derivative of position vs. time.
Do you know how to tell if the 2nd derivative of a function is negative, zero, or positive, just by looking at the graph of that function? I.e., without looking at the 1st derivative.
Mike Hunt
Sep7-08, 01:41 PM
Acceleration is the 2nd derivative of position vs. time.
Do you know how to tell if the 2nd derivative of a function is negative, zero, or positive, just by looking at the graph of that function? I.e., without looking at the 1st derivative.
no i dont. im in 11th grade high school.
Redbelly98
Sep7-08, 02:10 PM
Okay.
A good way to think about it then: is the slope of the graph increasing or decreasing at the given points? And when a=0, the slope is either constant, or the point is between a region of increasing slope and decreasing slopes.
What that means is:
For increasing acceleration, the graph of position curves upward (increasing velocity = increasing slope).
For decreasing acceleration, the graph curves downward.
For zero acceleration, the graph is neither curving upward nor downward.
Mike Hunt
Sep7-08, 03:09 PM
Okay.
A good way to think about it then: is the slope of the graph increasing or decreasing at the given points? And when a=0, the slope is either constant, or the point is between a region of increasing slope and decreasing slopes.
What that means is:
For increasing acceleration, the graph of position curves upward (increasing velocity = increasing slope).
For decreasing acceleration, the graph curves downward.
For zero acceleration, the graph is neither curving upward nor downward.
Yess thank you i got the first four boxes right. But now I'm having trouble with the last one finding where the speed decreases. I'm not sure what they mean.
http://i528.photobucket.com/albums/dd323/shazamanam/monk.jpg
Mike Hunt
Sep7-08, 03:54 PM
bump last problem! please!?
Redbelly98
Sep7-08, 04:58 PM
Speed = absolute value of velocity. They're asking if |v| is increasing or decreasing.
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