# Rectilinear Motion (Using Integrals)

• Apr 26th 2009, 06:30 PM
Tactified
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) = 3tē - 2t ; [the given information] s(0) = 1

Any help would be greatly appreciated, thanks!
• Apr 26th 2009, 06:37 PM
derfleurer
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.
• Apr 26th 2009, 06:48 PM
Tactified
Quote:

Originally Posted by derfleurer
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?
• Apr 26th 2009, 07:01 PM
Tactified
Nevremind about the constant, I just rememberd how to do it. So can anyone explain why I have to use an integral?
• Apr 26th 2009, 07:02 PM
derfleurer
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.

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

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.
• Apr 26th 2009, 07:12 PM
Tactified
Quote:

Originally Posted by derfleurer
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.