Given the position formula x = 3t² − 2t, what are the correct formulas for velocity and acceleration?

To determine the velocity and acceleration from the position formula x = 3t² − 2t, we need to apply differentiation with respect to time (t).

First, the velocity is the first derivative of the position with respect to time. So, if we differentiate the position function:

( v = \frac{dx}{dt} = \frac{d}{dt}(3t² - 2t) )

Applying the power rule of differentiation:

• The derivative of (3t²) is (6t).

• The derivative of (-2t) is (-2).

This gives us:

( v = 6t - 2 ).

Next, to find the acceleration, we differentiate the velocity function with respect to time:

( a = \frac{dv}{dt} = \frac{d}{dt}(6t - 2) ).

Using the power rule again:

• The derivative of (6t) is (6).

• The derivative of (-2) is (0).

Thus, we have:

( a = 6 ).

Therefore, the formulas for velocity and acceleration that we derived are:

• Velocity: ( v = 6