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Application of integration question help (car moving)

I'm unsure of why the position of the car will be given by the integral of v(t), my guess is that by using the limits and finding the are under the curve in the graph it will give the distance thus giving the position.

Question:
Assume that a car moves along a straight road with a velocity v(t), where v(t) is a known function of the time t. Explain why the position x(t) of the car after a time t=T will be given by the integral of v(t) between t=0 and t=T.
(assuming that x(0)=0 at time t=0)?

Here is my answer of how I think it works, I'm not too sure though and would like a longer answer showing better understanding but I can't really visualise it since there is x, v, and t, so I can't figure out a graph for it.

Answer: This will be given since by integrating with the limits t=T and t=0 it will work out the area under the curve.
Is my answer correct?
(edited 7 years ago)
Reply 1
Original post by Sayless
...


Simples. Velocity is the rate of change of position. So by the FToC, the position is the integral of velocity.
Reply 2
Original post by Zacken
Simples. Velocity is the rate of change of position. So by the FToC, the position is the integral of velocity.


Velocity is the rate of change of the car position therefore we can say v = dx/dt.By integrating v(t) in respect to t by using integral of vt dt, we will be left with dx, as it will cancel out the dt which is the position of the car, therefore shown.Is that a better answer?
Reply 3
Original post by Sayless
Velocity is the rate of change of the car position therefore we can say v = dx/dt.By integrating v(t) in respect to t by using integral of vt dt, we will be left with dx, as it will cancel out the dt which is the position of the car, therefore shown.Is that a better answer?


I think you mean this:

v=dxdtvdt=dxvdt=dx=xv = \frac{dx}{dt} \Rightarrow v \, \mathrm{d}t = \mathrm{d}x \Rightarrow \int v \, \mathrm{d}t = \int \, \mathrm{d}x = x.
Reply 4
Original post by Zacken
I think you mean this:

v=dxdtvdt=dxvdt=dx=xv = \frac{dx}{dt} \Rightarrow v \, \mathrm{d}t = \mathrm{d}x \Rightarrow \int v \, \mathrm{d}t = \int \, \mathrm{d}x = x.


Perfect, thanks

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