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A​ keyboarder's speed over a 6​-min interval is given by the following​ function, where W(t) is the​ speed, in words per​ minute, at time t.

W(t) = - 5 t^2 + 20 t + 60 ​
a) Find the speed at the beginning of the interval. ​
b) Find the maximum speed and when it occurs. ​
c) Find the average speed over the 6​-min interval.

User WISAM
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1 Answer

7 votes

Answer:

a) The speed at the beggining of the interval is 60 words per minute.

b) The maximum speed is 80 words per minute at the instant t = 2 minutes.

c) The average speed over the 6​-min interval is 60 words per minute.

Explanation:

a) Find the speed at the beginning of the interval. ​

This is W(0)


W(t) = -5t^(2) + 20t + 60

Then


W(0) = -5(0)^(2) + 20(0) + 60 = 60

The speed at the beggining of the interval is 60 words per minute.

b) Find the maximum speed and when it occurs. ​

Suppose we have a quadratic equation in the following format:


f(t) = at^(2) + bt + c

The vertex is the point


(t_(v), y(t_(v))

In which


t_(v) = -(b)/(2a)

In this question:


W(t) = -5t^(2) + 20t + 60

So


a = -5, b = 20, c = 60

Then


t_(v) = -(20)/(2*(-5)) = 2


W(2) = -5(2)^(2) + 20(2) + 60 = 80

The maximum speed is 80 words per minute at the instant t = 2 minutes.

c) Find the average speed over the 6​-min interval.

The average value of a function f(x) over an interval [a,b] is:


A = (1)/(b-a)\int\limits^b_a {x} \, dx

In this question:


A = (1)/(6-0)\int\limits^6_0 {W(t)} \, dt

So


F(t) = \int {W(t)} \, dt


A = (F(6) - F(0))/(6)

So


F(t) = \int {-5t^(2) + 20t + 60} \, dt


F(t) = -(5t^(3))/(3) + 10t^(2) + 60t


F(6) =  -(5*6^(3))/(3) + 10*6^(2) + 60*6 = 360


F(0) =  -(5*6^(0))/(3) + 10*0^(2) + 60*0 = 0

Then


A = (F(6) - F(0))/(6) = (360 - 0)/(6) = 60

The average speed over the 6​-min interval is 60 words per minute.

User AlexGreg
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