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Jessica Monique!

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*I have tried various ways to solve this problem and yet none seem to be right.. for those people who love physics here's a problem to tickle you brain with =) I appreciate all tries.. and if possible an explanation would be nice.. my head hurts from trying =/

Thanks



A particle moves along the x axis. Its position is given by the equation x = 1.70 + 3.00t - 3.90t^2 with x in meters and t in seconds.
(a) Determine its position when it changes direction.


(b) What is its velocity when it returns to the position it had at t = 0?
 
(a) The given equation is a parabolic equation. Therefore it will turn around at the vertex of the parabola. The easy way is simply to use the -B/2A rule (or by taking the derivative, same thing) to find the time at which the position changes.

t_turnaround = 3.9/(2*3) = .65sec

Plug this value back in to get the position.
1.7+3(.65)-3.9(.65)^2 = 2m

(b) It must have the same speed as when it began, though the velocity will be the opposite direction. Its initial velocity is +3m/sec (given by the coefficient on the middle term), so on the return it will have -3m/sec.
 
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