What are the correct formulas for the velocity and acceleration given the position function x = 2t^3 − 4t?

To determine the velocity and acceleration from the position function ( x = 2t^3 - 4t ), we need to find the first and second derivatives of the position function with respect to time ( t ).

The velocity ( v ) is defined as the first derivative of the position function:

[

v = \frac{dx}{dt}

]

Calculating the derivative of the position function, we differentiate ( 2t^3 - 4t ):

[

v = \frac{d}{dt}(2t^3) - \frac{d}{dt}(4t) = 6t^2 - 4

]

Next, the acceleration ( a ) is defined as the first derivative of the velocity, which is also the second derivative of the position:

[

a = \frac{dv}{dt}

]

Taking the derivative of the velocity function we just found:

[

a = \frac{d}{dt}(6t^2 - 4) = 12t

]

The derived formulas for velocity and acceleration are ( v = 6t^2 - 4 ) and ( a = 12t ), both of which match with