# Thread: differentiation and integration AGAIN

1. ## differentiation and integration AGAIN

hi, ive posted here before with regards to these subjects and im still struggling to grasp the concept. in my book theres an example,
an object moves along a straight line so that its position at time t in seconds is given by x=2t^3-6t (in meters) (t is greater than or equal to 0)

i)find the expresions for velocity and acceleration of the object at time t.

position - x=2t^3-6t
velocity - v=dx/dt=6t^2-6
acceleraration - a=dv/dt=12t

now where does the -6 come from in the velocity part :S and where does is go when you move onto the acceleration? thanks in advcnace...

2. Originally Posted by j55
hi, ive posted here before with regards to these subjects and im still struggling to grasp the concept. in my book theres an example,
an object moves along a straight line so that its position at time t in seconds is given by x=2t^3-6t (in meters) (t is greater than or equal to 0)

i)find the expresions for velocity and acceleration of the object at time t.

position - x=2t^3-6t
velocity - v=dx/dt=6t^2-6
acceleraration - a=dv/dt=12t

.
Derivative of -6t with respect to t is -6.
And derivative of a constant is zero.