Hi there!
Given the values for α and β, we can plug these into the given equation:
[tex]v(t) = 4t - 2t^2[/tex]
Begin by solving for the velocity at t = 0 and t = 3:
At t = 0:
[tex]v(0) = 4(0) - 2(0^2) = 0 \frac{m}{s}[/tex]
At t = 3:
[tex]v(3) = 4(3) - 2(3^2) = 12 - 18 = -6 \frac{m}{s}[/tex]
Acceleration is the SLOPE of the velocity graph (its derivative), so we can use the kinematic equation:
[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]
Plug in the knowns:
[tex]a = \frac{-6-0}{3 - 0} = \boxed{-2 \frac{m}{s^2}}[/tex]
