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

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