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

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

• A. v = 6t^2 − 4, a = 12t
• B. v = 2t^2 + 4, a = 4t
• C. v = 4t^2 + 2, a = 6t
• D. v = 12t − 4, a = 6t^2

Answer: (A)

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

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