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.
To find the average velocity of the ball during the first second, we need to calculate the change in position (height) of the ball divided by the change in time (seconds) during the first second.
The position of the ball is given by the equation: y = -16t^2 + 55t + 33
To find the average velocity during the first second, we need to calculate the change in position between t = 0 (start of the first second) and t = 1 (end of the first second).
Step 1: Substitute t = 0 into the equation to find the initial height of the ball:
y(0) = -16(0)^2 + 55(0) + 33
= 33
Step 2: Substitute t = 1 into the equation to find the final height of the ball after 1 second:
y(1) = -16(1)^2 + 55(1) + 33
= -16 + 55 + 33
= 72
Step 3: Calculate the change in position between t = 0 and t = 1:
Change in y = y(1) - y(0)
= 72 - 33
= 39
Step 4: Calculate the change in time (seconds) during the first second:
Change in t = 1 - 0
= 1
Step 5: Calculate the average velocity by dividing the change in position by the change in time:
Average velocity = Change in y / Change in t
= 39 / 1
= 39
Therefore, the average velocity of the ball during the first second is 39 feet per second.
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.