Calculas

sim0nz12345 · Sep 5, 2007, 05:34 AM

This is one hard question, could you please help me with it:

The position of a particle at any time, t , is represented by the anti-derivative of its velocity, v, with respect to time; that is, v=INTEGRATE (x dt). If the velocity is given the rule v=4t-5, and the initial position of the particle is 2cm left of the origin, find:

a) The rule for its position, x, at any time, t.
b) The position of the particle after 5 seconds.

The v=INTEGRATE (x dt), I had to write integrate because I don't have the sign for it but its being integrated with all the values in the brackets.

Thanks

Capuchin · Sep 5, 2007, 05:37 AM

I think you have a misunderstanding here.

$v = \frac{dx}{dt}$

$x = \int v dt$

To find the position function, you need to integrate the velocity function with respect to time.

sim0nz12345 · Sep 5, 2007, 05:41 AM

This seems more confusing to me
I would first need to integrate the velocity function which would be v=4t-5 with the time?

Capuchin · Sep 5, 2007, 05:45 AM

yup.

differential gives you the rate of change at a point (the gradient of a tangent, as your last question showed).

The velocity is the rate of change of the position-time graph, so the velocity is the derivative of position wrt time, so to get position back we have to integrate velocity wrt time.

You're right, you have to integrate v=4t-5 wrt time.

sim0nz12345 · Sep 5, 2007, 05:50 AM

So the integrated value of the velocity function would be v=2t^2-5t+c, where c is any constant number until solved.
And then how would I associate this with respect to time?
Thanks... seems to be challenging to me

Capuchin · Sep 5, 2007, 05:57 AM

it's not v anymore, it's x.

$v = \frac{dx}{dt}$

$v=4t-5$

$\frac{dx}{dt}=4t-5$

$dx=(4t-5)dt$

integrating:

$\int dx = \int (4t-5)dt$

$x = 2t^2-5t+c$

Does that make it clearer? I don't think it's strictly true but it makes it clearer right? :)

So now you have an equation for position in terms of time. You can work out c from this sentence in the question: "the initial position of the particle is 2cm left of the origin".

That will give you the answer to 1.

The answer for 2 can be easily solved from the answer to 1.

My work blocks this site in a few minutes so I apologise that I cannot give further answers for a few hours. I hope that this answer is comprehensive enough.

sim0nz12345 · Sep 5, 2007, 06:01 AM

Thank you, that's more than enough help for one day.
I really learnt a lot today thanks.

sim0nz12345 · Sep 5, 2007, 06:06 AM

Just to clarify something is the initial position of the particle is 2cm left of the origin, the coordinates (-2,0)
Thanks so much again

Capuchin · Sep 5, 2007, 08:47 AM

Erm, you would normally put time on the x axis and position on the y axis, so it's (0,-2) if you take left as negative. (the question doesn't seem to state whether "left" is positive or negative).

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