Thu October 27, 2011 By: Anagha Plus Two

The position of a particle is given by X(t) = 2t^2+3t^2-36t+40 after t seconds.The acceleration at time t is 12t+6.How did we arrive at the value 12t+6.PLS ELABORATE

Expert Reply
Sat October 29, 2011
In the given position the first term must be 2t3 and 2t2 that is:
x(t) = 2t3+3t2-36t+40
So the velocity = x'(t) = 6t2+6t-36
And the acceleration = x''(t) = 12t + 6
Related Questions
Mon September 05, 2016


Mon September 05, 2016

Root E^ root X, X>0.??

Ask the Expert