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cougar_21
Sep14-04, 12:22 PM
My problem reads:

A particle is observed to move with the coordinates x(t)=(1.5m/s)t + (-0.5 m/s^2)t^2 and y(t) = 6m + (-3m/s)t + (1.5 m/s^2)t^2. What are the particle's position, velocity, and acceleration? At what time(s) are the velocity's horizontal and vertical components equal?

I got the derivative of x(t) and y(t) to get v(x) and v(y). Where do I go from there :confused:

K.J.Healey
Sep14-04, 12:36 PM
do get the acceleration take the derivative of V(x) and V(y).

For the what time thing, set them equal, solve for "t"?

recon
Sep14-04, 09:21 PM
For the second question, solve for x(t) = y(t).