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Motion with Uniform Acceleration

by WannaLearn
Tags: acceleration, motion, uniform
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WannaLearn
#1
Mar27-10, 03:27 PM
P: 6
1. The problem statement, all variables and given/known data

In the experiment that I have performed, the ball was thrown upward and caught once it reached its initial position. The data was recorded using the graphical analysis program. The graph produced is a parabola. The parabolic curve was generated showing the values of three parameters: A:-4.453, B:9.835, C:-3.565.
I need to:
1). 1.Interpret the meaning of the fit coefficients A, B and C.
2). Find out the value of the acceleration due to gravity predicted by the experiment. Compare this to the expected value, and calculate the percentage difference.
3). Calculate the quantity −B/(2A), including units, and give a physical meaning to this quantity?

I am stuck, as I have not touched the physics book for more than 10 years. Any help will be appreciated.

2. Relevant equations

Y = A t2 + B t + C

3. The attempt at a solution

1). The coefficient A is acceleration, B - velocity, C -initial position of the ball?
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rock.freak667
#2
Mar27-10, 05:04 PM
HW Helper
P: 6,202
Yes that is correct.

For the second part, do you know a kinematic equation that resembles your formula of Y=At2+Bt+C?

For the third part, what do the units of B/A give?
WannaLearn
#3
Mar27-10, 05:42 PM
P: 6
Quote Quote by rock.freak667 View Post
Yes that is correct.

For the second part, do you know a kinematic equation that resembles your formula of Y=At2+Bt+C?

For the third part, what do the units of B/A give?

Thank you rock.freak,

2). The kinematic formula is x=x0+v0t+1/2at^2
3). I know that b/a would give me meters, and I assume by dividing -B/2A, I would get the highest point (the highest position the ball reaches). But when I look at the graph, it does not make sense to me. By dividing -B/2A I get 1.1m, while the graph shows the highest point of approx. 1.8m. Hope you can see the image I attached.

rock.freak667
#4
Mar27-10, 08:51 PM
HW Helper
P: 6,202
Motion with Uniform Acceleration

Quote Quote by WannaLearn View Post
2). The kinematic formula is x=x0+v0t+1/2at^2
3). I know that b/a would give me meters, and I assume by dividing -B/2A, I would get the highest point (the highest position the ball reaches). But when I look at the graph, it does not make sense to me. By dividing -B/2A I get 1.1m, while the graph shows the highest point of approx. 1.8m. Hope you can see the image I attached.
Quote Quote by WannaLearn View Post
The parabolic curve was generated showing the values of three parameters: A:-4.453, B:9.835, C:-3.565.
In your initial post, the values of A,B and C are shown above. These are not the same as the ones obtained by your plotting program.

How exactly did you get the first set of readings?
WannaLearn
#5
Mar27-10, 09:33 PM
P: 6
Quote Quote by rock.freak667 View Post
In your initial post, the values of A,B and C are shown above. These are not the same as the ones obtained by your plotting program.

How exactly did you get the first set of readings?
Sorry for the confusion. The values mentioned at the beginning of the post were from the first time I performed the curve fit, and in the image attached I performed curve fit for the second time. But even if I take the values from the image A:-4.764, B:10.51, C:-3.916, when I divide -B/2A, I get 1.1m, while on the graph the highest point is approx. 1.9m. Could you tell me what I am doing wrong?
rock.freak667
#6
Mar28-10, 12:07 AM
HW Helper
P: 6,202
Quote Quote by WannaLearn View Post
Sorry for the confusion. The values mentioned at the beginning of the post were from the first time I performed the curve fit, and in the image attached I performed curve fit for the second time. But even if I take the values from the image A:-4.764, B:10.51, C:-3.916, when I divide -B/2A, I get 1.1m, while on the graph the highest point is approx. 1.9m. Could you tell me what I am doing wrong?
According to the equation of your graph, it should give the same number. I am not sure why, it probably has to do with how the curve is being fitted to the points.
WannaLearn
#7
Mar28-10, 12:18 AM
P: 6
Quote Quote by rock.freak667 View Post
According to the equation of your graph, it should give the same number. I am not sure why, it probably has to do with how the curve is being fitted to the points.
Thank you rock.freak.

I appreciate your help. I am still stuck at part two. You refered to the kinematick equation x=x0+v0t+1/2at^2, is it to calculate the acceleration of the ball in motion? if yes, when I plug in the values for time and initial velocity, I get a -2.43 for acceleration. And that does not make sense to me.

I appreciate your patience, if you have any other suggestion.
rock.freak667
#8
Mar28-10, 12:28 AM
HW Helper
P: 6,202
Quote Quote by WannaLearn View Post
Thank you rock.freak.

I appreciate your help. I am still stuck at part two. You refered to the kinematick equation x=x0+v0t+1/2at^2, is it to calculate the acceleration of the ball in motion? if yes, when I plug in the values for time and initial velocity, I get a -2.43 for acceleration. And that does not make sense to me.

I appreciate your patience, if you have any other suggestion.
Remember from the first part, you said that A was acceleration right? (So the coefficient of t2 was the acceleration).

Thus similarly, if the coefficient of t2 in the kinematic equation is acceleration, then shouldn't they be the same?

i.e. -1/2g = A ? (negative since upwards is taken as the positive direction)


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