A particle is moving along the x-axis. At time t, the particle is at x(t)=2t^3-9t^2+12t+2. Find the acceleration of this particle at all time values where this particle stops.

Question

anonymous · Answer

differentiate it twice for the acceleration.
once for the velocity
and the other for the acc.

hartnn · Answer

velocity will be dx(t)/dt which u will get by once differentiating x(t)
acceleration is d^2x(t)/dt^2 which u will get by differentiating x(t) twice.

hartnn · Answer

and u already know the formula for d/dx (x^n), isn't it ?

anonymous · Answer

yes. For the first differentiation I got 6(x^2-3x+2), and for the second I got 6(2x-3).

hartnn · Answer

thats correct :) 
just put t instead of x.

hartnn · Answer

a(t) = 12t-18.

anonymous · Answer

yay! lol. thanks for your help!

hartnn · Answer

welcome :)