Question

pls only answer if you can actually help!!!

Answer 1

Answer:

p1. v(8) = (600 - 500) / (10 - 6) = 25 mmin-2, thus at that rate, in two minutes from t = 6, he gains 50 meters.

therefore v(8) = 550 mmin-1

p2. Final - Initial velocity over Final - Initial time

aavg = (-430 - 500) / (12 - 6) = - 155 mmin-2

p3. Take derivative of velocity function, and use the function to calculate acc at t = 3

dv(t) / dt = a = 1.5t2 + 4t

a(t=3) = 25.5mmin-2

p4. Δx/Δt = 300 - 200 / 3 - 0 = 100 / 3 = 33.33 mmin-1