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Need Some Math Help

A particle is moving along the x-axis so that at any time, t is greater than or equal to 0, its acceleration is given by a(t)= 18-2t. At time=1 the velocity of the particle is 36 meters per second and its position is x=21.

I found the velocity and the position function of the particle for when t is greater than or equal to 0 which is v(t)= 18t-t^2+19 and x(t)=9t^2+(t^3)/3+53/3.
I need help finding the position of the particle when it is farthest to the right but since t is just greater than or equal to 0, I don't really know how do so.

February 3, 2013

2 Comments • Newest first

Mathematician

[quote=2lazy2makeaname]set v(t)=0 and solve for when t=0, this will tell you when the particle changed direction
then plug those those values of t into x(t).[/quote]

Thank You!

Reply February 3, 2013
2lazy2makeaname

set v(t)=0 and solve for when t=0, this will tell you when the particle changed direction
then plug those those values of t into x(t).

Reply February 3, 2013