Answer:
The speed of rider is 8π m/min.
Explanation:
Given that
Radius = 10 m
Time t = 2 minutes
Distance = 16 m
According to figure,
Th height of rider at any time is (x+10) m.
From the right triangle
[tex]\sin\theta=\dfrac{x}{10}[/tex].....(I)
Differentiate both side with respect to t
[tex]\dfrac{d(sin\theta)}{dt}=\dfrac{1}{10}\times\dfrac{dx}{dt}[/tex]
Using chain rule,
[tex]\dfrac{d(sin\theta)}{d\theta}\dfrac{d\theta}{dt}=\dfrac{1}{10}\times\dfrac{dx}{dt}[/tex]
[tex]\cos\theta\dfrac{d\theta}{dt}=\dfrac{1}{10}\times\dfrac{dx}{dt}[/tex]
[tex]\sqrt{1-\sin^2\theta}\dfrac{d\theta}{dt}=\dfrac{1}{10}\dfrac{dx}{dt}[/tex]...(II)
When the height of the rider is 16 m.
[tex]h = x+10[/tex]
[tex]16-10=x[/tex]
Then the value of x is 6 m.
Put the value of x in the equation (I)
[tex]\sin\theta=\dfrac{6}{10}[/tex]
[tex]\sin\theta=\dfrac{3}{5}[/tex]
We need to calculate the value of [tex]\dfrac{d\theta}{dt}[/tex]
A Ferry wheel makes 1 revolution per 2 minutes.
[tex]\dfrac{d\theta}{dt}=\dfrac{2\pi}{2}[/tex]
[tex]\dfrac{d\theta}{dt}=\pi\ rad/m[/tex]
Now, substitute the value of [tex]\sin\theta[/tex] and [tex]\dfrac{d\theta}{dt}[/tex] in equation (II)
[tex]\sqrt{1-(\dfrac{3}{5})^2}\times\pi=\dfrac{1}{10}\dfrac{dx}{dt}[/tex]
[tex]\sqrt{1-\dfrac{9}{25}}\times\pi=\dfrac{1}{10}\dfrac{dx}{dt}[/tex]
[tex]\dfrac{dx}{dt}=8\pi\ meter/minute[/tex]
Hence, The speed of rider is 8π m/min.
