Question

2. An object’s velocity is measured to be v (t) = αt−βt2, where α = 4.00 m/s2 and β = 2.00 m/s3. Calculate the average acceleration of the object from t = 0.0 s to t = 3.0 s

Answer 1

Hi there!

Given the values for α and β, we can plug these into the given equation:

`v(t) = 4t - 2t^2`

Begin by solving for the velocity at t = 0 and t = 3:

At t = 0:

`v(0) = 4(0) - 2(0^2) = 0 (m)/(s)`

At t = 3:

`v(3) = 4(3) - 2(3^2) = 12 - 18 = -6 (m)/(s)`

Acceleration is the SLOPE of the velocity graph (its derivative), so we can use the kinematic equation:

`a = (v_f - v_i)/(\Delta t)`

Plug in the knowns:

`a = (-6-0)/(3 - 0) = \boxed{-2 (m)/(s^2)}`