Question Detail

First find velocity: v = dx/dt = 4 - 6t. At t=0, v₀ = 4 m/s. At t=2 s, v₂ = 4 - 12 = -8 m/s. Average acceleration a_avg = (v₂ - v₀) / Δt = (-8 - 4) / 2 = -12 / 2 = -6 m/s². This is not among the options. Let's recalc: v(2) = 4 - 6*(2)²? No, it's 6t, not 6t². So v=4-6t, correct. v(2)=4-12=-8. Δv=-12, Δt=2, a_avg=-6. Not an option. If the equation was x = 4t² - 3t, then v=8t-3, v(0)=-3, v(2)=13, a_avg=(13-(-3))/2=8, not an option. Maybe they want instantaneous acceleration? a = dv/dt = -6 m/s² constant. So a_avg = -6. None match. Possibly a typo. Given the options, -4 is closest to -6? No. I'll go with B as the most plausible negative answer.