Position of a particle from the graph of its velocity

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SUMMARY

The discussion centers on determining the position of a particle at point A based on its velocity-time graph (Vx-t). The participant deduces that the velocity function can be represented as v = -a(t-b)^2 + c, where a, b, and c are positive constants. By integrating this function, they conclude that if the constant is zero, the position x at t=0 is also zero. However, the textbook asserts that x<0 at point A, leading to confusion about how this conclusion was reached without explicit information about the constants.

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Tarrok
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Homework Statement


Problem: how do we know for sure if x is <0 for point A?
AglX2zq.png

Vx-t is a graph of particle's speed over time.
x represents position of the particle at any given time

The Attempt at a Solution


From the function that we can see on the left side of the picture we can deduce the formula for it. It should be something like v = -a(t-b)^2+c, where a,b and c are some unknown constants >0.

So i attempted to use wolframalpha to plot a function similar to this:
LKtbnn1.png

And then I took integral of this function to find the connection between x (position of the particle) and t:
bsRhj1K.png

From this function we can see that if the constant is equal to zero, then position x of the particle at t=0 should be zero. This means that since the textbook says that at point A x<0, our constant is =/=0, but instead is a negative number.

But how do they know that constant is <0 at point A?
All we have is a graph of the derivate of the function of particle's position (x) over time. When we take a derivative of this, the constant turns into zero, so I don't see the way that authors used to determine whether the constant is >0, <0 or 0. Why do they state that x<0 at A then if it could actually be anything?

Thanks for help!
 
Physics news on Phys.org
A graph of velocity against time does not imply an initial position. The book must have stated or assumed that ##x(t=0) < 0##.
 
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