Question

A boy moving along a circular path of radius 10m completes ${\dfrac{3}{4}^{th}}$ of the circle in $10s$. The magnitude of speed and velocity are
(A) $4.71m{s^{ - 1}}$ and $47.1m{s^{ - 1}}$
(B) $4.71m{s^{ - 1}}$ and $1.41m{s^{ - 1}}$
(C) $1.41m{s^{ - 1}}$ and $4.71m{s^{ - 1}}$
(D) $1.41m{s^{ - 1}}$ and $1.41m{s^{ - 1}}$

Answer 1

Hint We are given a circular path of a given radius and with a travel of a specific part of the circular trajectory and are asked to find the speed and velocity of the moving body. Thus, we will use the fundamental idea of the speed and velocity. In other words, we will use the formulation of the fundamental definition of speed and velocity of a body.

Formulae used:
$v = \dfrac{s}{t}$
Where,$v$ is the speed of the body, $s$ is the distance travelled by the body and $t$ is the time travelled by the body.
$v = \dfrac{s}{t}$
Where, $v$ is the velocity of the body, $s$ is the displacement of the body and $t$ is the time travelled by the body.

Complete step by step solution:
Here,
We are given that the boy completes${\dfrac{3}{4}^{th}}$ of the circle.
Thus,
The distance travelled by the boy,${s_d} = \dfrac{3}{4} \times 2\pi \left( {10} \right)$
Further, we get
${s_d} = 15\pi $
Then, we get
${s_d} = 47.12m$
Now,
The time of travel,$t = 10s$
Thus,
Speed of the body,${v_s} = \dfrac{{47.12}}{{10}}$
Further, we get
${v_s} = 4.71m{s^{ - 1}}$
Now,
The displacement of the boy,$s = 10\sqrt 2 m$
Further, we get
$s = 14.14m$
Thus,
The velocity of the boy,$v = \dfrac{{14.1}}{{10}}$
Hence,
$v = 1.41m{s^{ - 1}}$

Hence, the correct answer is (2).

Note We are evaluating the answer by using the fundamental definition of speed and velocity. The external factors such as the friction and other parameters remained negligible throughout the evaluation of the solution. If in case, the friction acted on the body, then the velocity and the speed would have been something different.