Answer:
Robbie Glider
Step-by-step explanation:
Given
Melissa Glider
[tex]m(s) = 0.4(s^3 - 11s^2 + 31s - 1)[/tex]
Robbie Glider
See attachment for function
Required
Which reaches the greater maximum within the first 6 seconds
Melissa Glider
First, we calculate the maximum of Melissa's glider
[tex]m(s) = 0.4(s^3 - 11s^2 + 31s - 1)[/tex]
Differentiate:
[tex]m'(s) = 0.4(3s^2 - 22s + 31)[/tex]
Equate to 0 to find the maximum
[tex]0.4(3s^2 - 22s + 31) = 0[/tex]
Divide through by 0.4
[tex]3s^2 - 22s + 31 = 0[/tex]
Solve for s using quadratic formula:
[tex]s = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}[/tex]
Where
[tex]a = 3; b = -22; c = 31[/tex]
So:
[tex]s = \frac{22 \± \sqrt{(-22)^2 - 4*3*31}}{2*3}[/tex]
[tex]s = \frac{22 \± \sqrt{112}}{6}[/tex]
[tex]s = \frac{22 \± 10.6}{6}[/tex]
Split:
[tex]s = \frac{22 + 10.6}{6}\ or\ s = \frac{22 - 10.6}{6}[/tex]
[tex]s = \frac{32.6}{6}\ or\ s = \frac{11.4}{6}[/tex]
[tex]s = 5.4\ or\ s = 1.9[/tex]
This implies that Melissa's glider reaches the maximum at 5.4 seconds or 1.9 seconds.
Both time are less than 6 seconds
Substitute 5.4 and 1.9 for s in [tex]m(s) = 0.4(s^3 - 11s^2 + 31s - 1)[/tex] to get the maximum
[tex]m(5.4) = 0.4(5.4^3 - 11*5.4^2 + 31*5.4 - 1)[/tex]
[tex]m(5.4) = 1.24ft[/tex]
[tex]m(5.4) = 0.4(1.9^3 - 11*1.9^2 + 31*1.9- 1)[/tex]
[tex]m(5.4) = 10.02ft[/tex]
The maximum is 10.02ft for Melissa's glider
Robbie Glider
From the attached graph, within an interval less than 6 seconds, the maximum altitude is at 3 seconds
[tex]r(3) = 22ft[/tex]
Compare both maximum altitudes, 22ft > 10.02ft. This implies that Robbie reached a greater altitude
