Question

the position of an object is given by: x(t)=4t^2+8t+12 what is the acceleration of the object at t=250 seconds?

Answer 1

We will have the following:

First, we remember that velocity is the change of position with respect to time and this can be seen as the derivative of the position; and that the acceleration is the change of velocity with respect to time and this can be seen as the derivative of the velocity or the second derivative of the position; so:

`x(t)=4t^2+8t+12`

Then velocity will be:

`v(t)=8t+8`

And finally the acceleration will be:

`a(t)=8`

And thus, when we examine the problem at t = 250s we will have that the acceleration will be 8 units of distance / s^2; this due to the fact that the second derivative is a constant.