9514 1404 393
Answer:
-10 m/s²
Step-by-step explanation:
The acceleration is the rate of change of velocity. The average rate of change of velocity will be the difference over the interval, divided by the width of the interval.
(V(2) -V(0))/(2 -0) = (40 -5(2²)) -(40 -5(0²)) = (20 -40)/2 = -10 . . . m/s²
The average acceleration on [0, 2] is -10 m/s².
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Additional comment
The average rate of change of a quadratic function on an interval is exactly equal to the derivative of the function at the midpoint of the interval:
Vx'(t) = -10t ⇒ Vx'(1) = -10 . . . m/s² (midpoint of [0, 2] is t=1)