# Find the Acceleration Given a Position Function| University Physics

### An object moves along the x-axis. The position as a function of time is given by $x\left(t\right)=at^3+bt+c$ with$a=4.00\:m/s^3$$a=2.00\:m/s$$a=1.00\:m/s$What is the object’s acceleration at time t = 0.50 s?

ANSWER: $12.00\:m/s^2$

So, we are given a position function $x\left(t\right)=4t^3+2t+1$

The velocity function can be determined by getting the derivative of the position function. That is

$v\left(t\right)=\frac{d}{dt}\left(x\left(t\right)\right)$

$v\left(t\right)=\frac{d}{dt}\left(4t^3+2t+1\right)$

$v\left(t\right)=12t^2+2$

The acceleration function is the derivative of the velocity function. That is

$a\left(t\right)=\frac{d}{dt}\left(v\left(t\right)\right)$

$a\left(t\right)=\frac{d}{dt}\left(12t^2+2\right)$

$a\left(t\right)=24t$

The acceleration of the object at time t=0.50 s is

$a\left(0.50\right)=24\left(0.5\right)=12.00\:m/s^2$