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Question: An athlete runs on a circular track, whose radius is 50 m with a constant speed. It takes 50 second ...

An athlete runs on a circular track, whose radius is 50 m with a constant speed. It takes 50 second to reach a diametrically opposite point B from starting point A. Find:
(a) Distance covered
(b) Displacement and
(c) Speed.

Explanation

Solution

Hint: i) Moving between point A and B, the athlete covered half of the circumference of the circular track which would determine the distance covered by him.
ii) In case of circle, diameter is possible shortest path travelled between diametrically opposite points.
Formula used
2πr2\pi r= circumference of circular track
speed=distance coveredtime taken\Rightarrow speed = \dfrac{distance\ covered}{time\ taken}

Complete Step by Step solution
Given, Athletes run with a constant speed. That means, he covers the same distance at the same time. For instance, if he covers 3 meters in 10 seconds then in another 10 seconds he covered 3 meters again.

To find distance covered by him while running on a circular track from point A to B.
So, the distance covered by him in 50 second = trackcircumference2\dfrac{{track circumference}}{2} because, distance covered will become equal to half the circumference of the circle.
Also, given that, radius of circular track = 50 m
distancecovered=2πr2\Rightarrow dis\tan ce\operatorname{cov} ered = \dfrac{{2\pi r}}{2}…………………………………..eq.1
Where 2πr2\pi r= circumference of circular track
And r = radius of track
It’s clear from figure, that radius of circular track = 50 m, so circumference is given by:
2πr=2×3.14×50\Rightarrow 2\pi r = 2 \times 3.14 \times 50
2πr=314.15\Rightarrow 2\pi r = 314.15……………………………………eq.2
Substituting values from eq.2 in eq.1, we get,

distance operatornamecovered=314.152 distance operatornamecovered=157.08meter  \Rightarrow dis\tan ce\ operatorname{cov} ered = \dfrac{{314.15}}{2} \\\ \Rightarrow dis\tan ce\ operatorname{cov} ered = 157.08meter \\\

\therefore The total distance covered by an athlete to reach a diametrically opposite point B from starting point A is 157.08 meter.

(B) To find displacement :
Now, displacement of a particle is the change in the position of a particle in a particular direction.
Or in simple words, it is the shortest path that should be covered by an object between two fixed points.
See in below given figure, the shortest path between points A and B is shown as:

So, moving on a straight path from A to B gives us displacement between these points and this path is the diameter of a circular running track.
\Rightarrow Displacement = diameter of circular track
displacement=2×radius\Rightarrow displacement = 2 \times radius

displacement=2×50 displacement=100meter  \Rightarrow displacement = 2 \times 50 \\\ \Rightarrow displacement = 100meter \\\

Since, displacement is a vector quantity and hence always associated with a particular direction. So, the direction of displacement would be from A to B.
Therefore, displacement between point A and B is of magnitude 100 meter with a direction from A to B.
(C) To find the speed of athlete:
Since, in kinematics we all have studied the time rate of change of position of an object is called speed. Or, the ratio of total distance covered to total time covered.
speed=distancecoveredtime taken\Rightarrow speed = \dfrac{{dis\tan ce\operatorname{cov} ered}}{{time\ taken}}
speed=track circumference2×1time taken\Rightarrow speed = \dfrac{{track\ circumference}}{2} \times \dfrac{1}{{time\ taken}}
speed=2πr2×1time taken\Rightarrow speed = \dfrac{{2\pi r}}{2} \times \dfrac{1}{{time\ taken}}
It takes 50 second to reach a diametrically opposite point B from starting point A.
\RightarrowTime taken = 50 seconds …………………eq.3
\therefore Substituting the values from eq.1, eq.2 and eq.3 we get,
speed=314.152×150\Rightarrow speed = \dfrac{{314.15}}{2} \times \dfrac{1}{{50}}
speed=314.15100\Rightarrow speed = \dfrac{{314.15}}{{100}}
speed=3.1415\Rightarrow speed = 3.1415meter /sec or,
speed=3.1415\Rightarrow speed = 3.1415m /sec or,
speed=3.1415ms1\Rightarrow speed = 3.1415m{s^{ - 1}}

Note It should be noted that the actual distance travelled by an object is either equal or greater than magnitude of displacement because displacement is the shortest path covered by object. Also, we above calculated that distance = 157.08 meter and displacement = 100 meter. And always displacement is associated with direction, so determine direction always along with magnitude of displacement and velocity.