Answer:
23. 0.4583 seconds
24. 0.0107 seconds
Step-by-step explanation:
The problem statement tells you how to work it. You need to convert speed from miles per hour to feet (or inches) per second.
90 mi/h = (90·5280 ft)/(3600 s) = 132 ft/s = (132·12 in)/s = 1584 in/s
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23. The time it takes for the ball to travel 60.5 ft is ...
time = distance/speed
time = (60.5 ft)/(132 ft/s) = 0.4583 s
It takes 458.3 milliseconds to reach home plate.
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24. time = distance/speed
time = (17 in)/(1584 in/s) = 0.0107 s
The ball is in the strike zone for 10.7 milliseconds.
(a) The "average value" of a function over an interval [a,b] is defined to be
(1/(b-a)) times the integral of f from the limits x= a to x = b.
Now S = 200(5 - 9/(2+t))
The average value of S during the first year (from t = 0 months to t = 12 months) is then:
(1/12) times the integral of 200(5 - 9/(2+t)) from t = 0 to t = 12
or 200/12 times the integral of (5 - 9/(2+t)) from t= 0 to t = 12
This equals 200/12 * (5t -9ln(2+t))
Evaluating this with the limits t= 0 to t = 12 gives:
708.113 units., which is the average value of S(t) during the first year.
(b). We need to find S'(t), and then equate this with the average value.
Now S'(t) = 1800/(t+2)^2
So you're left with solving 1800/(t+2)^2 = 708.113
<span>I'll leave that to you</span>
I think the answer is maximum