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Old April 26th, 2009, 08:30 PM
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Default Rectilinear Motion (Using Integrals)

Ok, i'm a bit confused on what this question is asking. I dont know how to visualize it. Here's the question:

A particle moves along an s-axis. Use the given information to find the position function of the particle.

V(t) = 3t² - 2t ; [the given information] s(0) = 1

Any help would be greatly appreciated, thanks!
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Old April 26th, 2009, 08:37 PM
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What's the s-axis? And s(0)? If s is your time in seconds, is the function s returning time based on acceleration, distance, velocity, what?

I'm assuming your distance at time 0 is 1. In which case you just take the integral of V(t) and your constant of integration becomes 1.

D(t) = x^3 + x^2 + 1

Edit: Well this seems to explain it better than my feeble mind.
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Old April 26th, 2009, 08:48 PM
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Quote:
Originally Posted by derfleurer View Post
What's the s-axis? And s(0)? If s is your time in seconds, is the function s returning time based on acceleration, distance, velocity, what?

I'm assuming your distance at time 0 is 1. In which case you just take the integral of V(t) and your constant of integration becomes 1.

D(t) = x^3 + x^2 + 1

Well this seems to explain it better than my feeble mind.
I'm not sure what the s-axis is, I think its position. Time doesnt matter. I think the function is based off V(t), in which case for what i'm doing is velocity.

I guess I dont understand why i'm taking the integral. The answer in the solutions part of the book gives the same function you described. Also how do you know 1 is the constant?
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Old April 26th, 2009, 09:01 PM
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Nevremind about the constant, I just rememberd how to do it. So can anyone explain why I have to use an integral?
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Old April 26th, 2009, 09:02 PM
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According to that page, s(t) is your distance. And at t = 0, s = 1.

V(t) is a velocity-time graph. When you integrate V over a span of dt, you essentially get v * s (or m/s * s), which is just m.

\int V(t)dt = s(t)

The integral yeilds s(t) = x^3 + x^2 + c. And again, at t = 0, s = 1.

1 = (0)^3 + (0)^2 + c

Thus c = 1.
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Old April 26th, 2009, 09:12 PM
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Quote:
Originally Posted by derfleurer View Post
According to that page, s(t) is your distance. And at t = 0, s = 1.

V(t) is a velocity-time graph. When you integrate V over a span of dt, you essentially get v * s (or m/s * s), which is just m.
OOOOooooohhhh. Yeah alright, I forgot that velocity isnt just a single variable. That helps a lot, thanks.
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