A ball is launched straight up in the air from a height of 7 feet. Its velocity (feet/second) $t$ seconds after launch is given by $f(t)=-32 t+289$. What is the ball's average velocity from 1 seconds to 9 seconds?
Between 1 seconds and 9 seconds the ball's average velocity is $\square$ feet/second.
If necessary, round the answer to two decimal places.
Answer
So, the ball's average velocity from 1 second to 9 seconds is \(\boxed{-32 \text{ feet/second}}\). This negative sign indicates that the ball is moving downwards on average over this time period.
Step 1 ：The average velocity of a function over an interval [a, b] is given by the formula: \(\text{Average velocity} = \frac{f(b) - f(a)}{b - a}\)
Step 2 ：Here, a = 1 second and b = 9 seconds. The function f(t) = -32t + 289 gives the velocity at any time t.
Step 3 ：Substitute these values into the formula: \(\text{Average velocity} = \frac{f(9) - f(1)}{9 - 1}\)
Step 4 ：Calculate the values: \(\text{Average velocity} = \frac{(-32*9 + 289) - (-32*1 + 289)}{8}\)
Step 5 ：Simplify the equation: \(\text{Average velocity} = \frac{(1 - 257)}{8}\)
Step 6 ：Calculate the final value: \(\text{Average velocity} = -32 \text{ feet/second}\)
Step 7 ：So, the ball's average velocity from 1 second to 9 seconds is \(\boxed{-32 \text{ feet/second}}\). This negative sign indicates that the ball is moving downwards on average over this time period.
