Question

A person with his hands in his pocket is skating on ice at the rate of \[10m{s^{ - 1}}\]and describes a circle of radius of 50m. What is his inclination to the vertical? \[\left( {g = 10m{s^{ - 2}}} \right)\]
A. \[{\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]
B. \[{\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right)\]
C. \[{\tan ^{ - 1}}\left( {\dfrac{3}{5}} \right)\]
D. \[{\tan ^{ - 1}}\left( {\dfrac{1}{{10}}} \right)\]

Answer 1

Hint:In order to solve this problem we need to know what data they have provided. They have given the velocity and radius of a circle. Using this data, we are going to find the angle of inclination as shown below.

Formula Used:
To find the inclination of a person from vertical is given by the formula,
\[\tan \theta = \dfrac{{{v^2}}}{{rg}}\]
Where, v is velocity, r is radius and g is acceleration due to gravity.

Complete step by step solution:
When a person with his hands in his pocket is skating on ice at the rate of \[10\,m{s^{ - 1}}\] and describes a circle of radius 50 m. We need to find his inclination to the vertical that is \[\theta \]. To find the inclination of the person from vertical is given by the formula,
\[\tan \theta = \dfrac{{{v^2}}}{{rg}}\]
Here, \[v = 10m{s^{ - 1}}\], \[r = 50m\] and \[g = 10m{s^{ - 2}}\]

Substitute the values in the above equation we get,
\[\tan \theta = \dfrac{{{{\left( {10} \right)}^2}}}{{50 \times 10}}\]
\[\Rightarrow \tan \theta = \dfrac{1}{5}\]
\[\therefore \theta = {\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right)\]
Therefore, his inclination to the vertical is \[{\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right)\].

Hence, option B is the correct answer.

Note: Here in the given problem it is important to remember that the equation for the angle of inclination to the vertical from that using the given values we can easily find the solution. The inclination of an angle or a line is the acute or obtuse angle that is formed when a non-horizontal line intersects the x-axis. The slope of a line is the tangent of the angle of inclination. It is denoted by m. The slope can be positive, negative, zero, or even undefined.