avg and instantaneous accel question

by missrikku
Tags: accel, instantaneous
Sep17-03, 07:43 PM
P: n/a
hello again,

I am having trouble with this problem:

A particle leaves the origin with an initial velocity v = (3.00i) m/s and a constant acceleration a = (-1.00i - 0.500j) m/s^2. When the particle reaches its max x coordinate, what are a)it's velocity and b) its position vector.

Am I to assume that at the max the v = 0 m/s? That means I've got to assume that the motion is a parabola, but I don't think that I have enough information to assume that. Or do I? I would just like to know how to start this problem out. Can someone lead me in the right direction? Thanks!
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FZ+ is offline
Sep17-03, 07:47 PM
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P: 1,954
I think considering at max x, the x component of v (but not neccessarily the y!) to be zero is pretty reasonable, as you are looking for turning points.
HallsofIvy is offline
Sep18-03, 05:13 AM
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PF Gold
P: 38,900
If v> 0 then the particle is still moving to the right: it's not yet at it's maximum x value.

If v< 0 then is already moving to the left: it's coming back from it's maximum x value.

In either case, the particle is not AT it's maximum x value.

In order to be AT it's maximum x value, the particle's speed MUST be 0- that's not an "assumption"!

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