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Posted: 1/13/2002 5:36:35 AM EDT
I am doing a math project. a quadratic equation is written as: y = ax2(should be superscript)+bx+c anyway I thought if it involved gravity the value for A would become -9.8 but my math book says -4.9, why is this?
Link Posted: 1/13/2002 5:46:03 AM EDT
I believe you are thinking of velocity as a function of time due to gravity. It is actually a definite integral of "a" (acceleration) over the interval from zero to "t" (time), and works out to be: 1/2 a t-squared Since "a" is 9.8 m/sec-squared, "1/2 a" would be 4.9 m/sec-squared. This is a guess, though, since you are quoting the math book out of context.
Link Posted: 1/13/2002 5:48:22 AM EDT
Link Posted: 1/13/2002 6:27:53 AM EDT
Originally Posted By CanadianGunNut: I am doing a math project. a quadratic equation is written as: y = ax2(should be superscript)+bx+c anyway I thought if it involved gravity the value for A would become -9.8 but my math book says -4.9, why is this?
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What school are YOU going to?! Sheeesh! Obviously you're missing the most important issue here. Namely, [b]how does this equation make you feel?[/b] Do you feel like you are being oppressed or that your self-esteem is hurt? Does this assignment, or algebra in general, cause you any shame? Are you aware of any racist undertones in the question. Please get in touch with your true feelings about this - you probably are being discriminated against and shamed and yet you're so programmed to "just get the answer right" that you don't even see the greater insult and damage being done to you. You need to explore the cultural bias inherent in the question. I'm surprised you can't see damage being done to you're self-esteem by the racist/sexist/homophobic implications of you're assignment. You DESPERATELY need some sensitivity-training, cultural-bias-awareness, diversity&multicultural-awareness training which is FAR more important to your education than "math".
Link Posted: 1/13/2002 6:53:28 AM EDT
Um...I thought the quadratic equation was A(squared)+B(squared)=c{squared}? It is the relationship of the three sides of a triangle where one angle is 90degrees,c being the longest side. Once you have the lenghts of the sides figured out you can then resolves the angles since you already know that one is 90 degrees and the total sum of angles if 180 degrees. Sgtar15
Link Posted: 1/13/2002 7:03:43 AM EDT
Originally Posted By sgtar15: Um...I thought the quadratic equation was A(squared)+B(squared)=c{squared}? It is the relationship of the three sides of a triangle where one angle is 90degrees,c being the longest side. Once you have the lenghts of the sides figured out you can then resolves the angles since you already know that one is 90 degrees and the total sum of angles if 180 degrees. Sgtar15
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That's the Pythagorean Theorem.
Link Posted: 1/13/2002 7:16:08 AM EDT
[Last Edit: 1/13/2002 7:18:02 AM EDT by e8ght]
Sounds like position as a function of time in the vertical plane for an object being acted upon by a constant acceleration (ie gravity). -9.8M/s^2 is the force of gravity on earth, give or take. '-4.9' (UNITS!! - calculating and including the full units in your equations allows you to double check for any errors) could be '1/2a' as marvl points out. Or it could be 'a' on another planet, a favorite topic for some math professors who seldom get invited places... :) The full equation and its derivation is at [URL]http://stravinsky.ucsc.edu/~josh/5A/book/notes/node22.html[/URL] Edited to make link active
Link Posted: 1/13/2002 7:25:25 AM EDT
Originally Posted By marvl: I believe you are thinking of velocity as a function of time due to gravity. It is actually a definite integral of "a" (acceleration) over the interval from zero to "t" (time), and works out to be: 1/2 a t-squared Since "a" is 9.8 m/sec-squared, "1/2 a" would be 4.9 m/sec-squared. This is a guess, though, since you are quoting the math book out of context.
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I have math anxiety... [puke]
Link Posted: 1/13/2002 8:00:07 AM EDT
Ouch: Now my head hurts !!! Jay [>:/]
Link Posted: 1/13/2002 8:33:53 AM EDT
Here's my weak, half-asleep effort. Since you mention gravity and quadratic equations, it sounds like you are studying parabolic trajectories. The quadratic equation will describe the trajectory, while the downward acceleration of gravity will be controlled by the -9.81m/s^2. So, r = r(initial) + v*t - 1/2 g t^2 The 2 denominator in the above equation accounts for your question of why the acceleration of gravity is half the value you think it ought to be. "Motion with Constant Acceleration", from Resnick/Halliday/Krane
Link Posted: 1/13/2002 8:53:05 AM EDT
Damn math teachers practicing physics without a license. Why do they insist on confusing things by not using units, using the letter x to stand for time instead of t, and using a to stand for 1/2 acceleration... Anyway y=ax^2+bx+c y=position as function of time x=time in seconds a=1/2 acceleraton=-4.9meters/sec^2 b=velocity at time 0 c=position at time 0 For example, you're standing on the roof at a height of 4 meters on New Years Eve. You shoot your AR straight up in the air and it has a muzzle velocity of 900meters/sec. How high will the bullet be after one second? a=-4.9, b=900, c=4, x=1, solve for y y=-4.9*1+ 900*1 + 4 = 899.1 meters.
Link Posted: 1/13/2002 9:03:59 AM EDT
Originally Posted By marvl: I believe you are thinking of velocity as a function of time due to gravity. It is actually a definite integral of "a" (acceleration) over the interval from zero to "t" (time), and works out to be: 1/2 a t-squared Since "a" is 9.8 m/sec-squared, "1/2 a" would be 4.9 m/sec-squared. This is a guess, though, since you are quoting the math book out of context.
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Good job marvl - this seems to be the most logical assessment. Integration is the only way to explain the value alteration for "a". I'd assume that the formula is a function of acceleration, and the goal is to find velocity, which is = to the area under the acceleration curve = the integration of the acceleration formula over the interval in question ([a, b], generically).
Link Posted: 1/13/2002 9:29:46 AM EDT
Next time Lockheed Martin does some engineering work on a space shuttle, I'm gonna recommend you fellas! [:E]
Link Posted: 1/13/2002 10:37:18 AM EDT
Great idea ECS, that way we can finally get that poor old VentureStar back on track. It's gonna take more than familiarity with the quadratic formula to solve its problems :)
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