A particle is moving on x

[sammy87]

A particle is moving along the x-axis with position function x(t). Find a) velocity and b) acceleration, and c) describe the motion of the particle for t is greater than or equal to 0.

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x(t) = 6 - 2t - t²

a) velocity = dx/dt

x'(t) = 0 - 2 - 2t

x'(t) = -2t - 2

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b) acceleration = dx/dt<sup>2

x''(t) = -2</sup>

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^{c) ??}

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I have no idea how the particle is moving. Since time cannot be negative, I looked at the velocity graph and started at t = 0. Since -2t - 2 is a linear line, and it is always negative after t = 0, I think that the particle is going backwards after starting at x = 2? Also, is the particle going 2 units/sec²? That is all I can think of, and I know I'm wrong. Help?

 

[galactus]

Are you sure that is not \(\displaystyle -t^{2}+2t+6\)

That would make more sense. That way, the particle would be stopped at x=7 when t=1 second.

The way it is now, you are correct. It is stopped at t = -1 second.

It is at x=6 when t=0 and moves in the negative direction while decelerating at -2.