Aparticle is moving in a plane with constant radial velocity = 2.11 m/s, having started at the origin. it also has a constant angular velocity = 2.55 rad/s. when the particle is 2.64 m from the origin, what is the magnitude of its velocity in m/s?

Magnitude of velocity =14.20m/s

Explanation:

The equation for the magnitude of velocity is given by:

/V/= SQRT r^2 +r^2×angular velocity ^2

/V/ = SQRT 2.11^2+ 2.64^2+2.55^2

/V/= SQRT 4.4521 + (6.9696 ×6.5025)

/V/=SQRT 201.768

/V/=14.20m/s

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 .

answera compound;