posted by Jackie on .
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.
the average velocity will be distance traveled divided by the time interval
y(0) = 33
y(1) = 72
avg v = 72-33 = 39
v = -32t + 55
avg is the ∫[0,1]v(t)/(1-0)
Note that the value is just that figured above.