Respuesta :
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.
Step-by-step explanation:
a) Find the speed at the beginning of the interval.
This is W(0)
[tex]W(t) = -5t^{2} + 20t + 60[/tex]
Then
[tex]W(0) = -5(0)^{2} + 20(0) + 60 = 60[/tex]
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:
[tex]f(t) = at^{2} + bt + c[/tex]
The vertex is the point
[tex](t_{v}, y(t_{v})[/tex]
In which
[tex]t_{v} = -\frac{b}{2a}[/tex]
In this question:
[tex]W(t) = -5t^{2} + 20t + 60[/tex]
So
[tex]a = -5, b = 20, c = 60[/tex]
Then
[tex]t_{v} = -\frac{20}{2*(-5)} = 2[/tex]
[tex]W(2) = -5(2)^{2} + 20(2) + 60 = 80[/tex]
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:
[tex]A = \frac{1}{b-a}\int\limits^b_a {x} \, dx[/tex]
In this question:
[tex]A = \frac{1}{6-0}\int\limits^6_0 {W(t)} \, dt[/tex]
So
[tex]F(t) = \int {W(t)} \, dt[/tex]
[tex]A = \frac{F(6) - F(0)}{6}[/tex]
So
[tex]F(t) = \int {-5t^{2} + 20t + 60} \, dt[/tex]
[tex]F(t) = -\frac{5t^{3}}{3} + 10t^{2} + 60t[/tex]
[tex]F(6) = -\frac{5*6^{3}}{3} + 10*6^{2} + 60*6 = 360[/tex]
[tex]F(0) = -\frac{5*6^{0}}{3} + 10*0^{2} + 60*0 = 0[/tex]
Then
[tex]A = \frac{F(6) - F(0)}{6} = \frac{360 - 0}{6} = 60[/tex]
The average speed over the 6-min interval is 60 words per minute.
