Question

A ball is shot into the air from a bridge, and its height, y, in feet, above the ground t seconds after it is thrown is given by y=f(t) = −16t2 + 55t + 33.
What is the average velocity of the ball during the first second? Round your answer to two decimal places.

Answer 1

the average velocity will be distance traveled divided by the time interval

y(0) = 33
y(1) = 72
avg v = 72-33 = 39

Or,
v = -32t + 55
avg is the ∫[0,1]v(t)/(1-0)

Note that the value is just that figured above.