While her kid brother is on a wooden horse at the edge of a merry-go-round, Sheila rides her bicycle parallel to its edge. The wooden horses have a tangential speed of 6 m/s. Sheila rides at 4 m/s. The radius of the merry-go-round is 8 m. At what time intervals does Sheila encounter her brother, if she rides opposite to the direction of rotation of the merry-go-round? a. 5.03 s
b. 8.37 s
c. 12.6 s
d. 25.1 s
e. 50.2 s

t = 5.03 s

Explanation:

To find the time interval when Sheila encounter her brother, you first calculate the angular speed of both Sheila and her brother.

You use the following formula:

w: angular speed

v: tangential speed

r: radius of the trajectory = 8 m

For  you have:

For her brother:

Next, they will encounter to each other when the angular distance of the Brother of sheila equals the angular distance of Sheila in the opposite direction. This can be written as follow:

They encounter for θ = 2π-θ':

You replace the values of the parameters in the previous equation and solve for t:

Hence, Sheila encounter her brother in 5.03 s

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we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .