A keyboarders speed over a 6min interval is given by the fol

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) = -4t^2 + 16t + 48 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. a) The speed at the beginning of the interval is words per minute. (Simplify your answer.)

Solution

w = -4t^2 + 16t + 48

a) Beginning :
t = 0
w = 0 + 0 + 48
So, at the beginning, the speed is 48 wpm ---> ANSWER

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b)
MAx speed :
w\' = -8t + 16 = 0
8t = 16
t = 2

Lets confirm that this indeed is the time of max speed
by finding the second derivative

w\'\' = -8, which is negative,
and thus the critical is a MAX

When t = 2, we have :
w = -4(2)^2 + 16(2) + 48
w = -16 + 32+ 48
w = 64 wpm

So, max speed = 64 wpm
and it occurs when t =2 ----> ANSWER

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c)
For this, we just have to use the formula

1/(b-a) * (integral from a to b) w(t) dt

1/(6-0) * (int from 0 to 6) (-4t^2 + 16t + 48) dt

Solving, we get :
48

So, the avg speed = 48 wpm