Question

An athlete is running on a circular track. He runs a distance of 100 m in 5 s before

returning to his original position. What is his average speed and velocity?​

Answer 1

Answer:

He runs a distance of 100 m in 5 second before returning to his original position. ~ As it's given that He runs a distance of 100 m in 5 second "before returning to his original position" Henceforth, the displacement be 0 as because he returned to his original position.

Step-by-step explanation:

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Answer 2

Answer:

${\huge{\sf{\underline{Required \: solution...}}}}$

${\bigstar{\sf{\underline{About \: question...}}}}$

⋆ Correct question: An athlete is running on a circular track. He runs a distance of 100 m in 5 s before returning to his original position. What is his average speed and velocity?

⋆ Understanding the question: This question says that we have to find out the average speed and velocity of an athlete who is running on a circular track. He runs a distance of 100 metres in 5 seconds before returning to his original position.

${\bigstar \:{\pmb{\sf{\underline{Given \: that...}}}}}$

⋆ An athlete is running on a circular track.

⋆ He runs a distance of 100 m in 5 second before returning to his original position.

${\bigstar \:{\pmb{\sf{\underline{To \: find...}}}}}$

⋆ The average speed of athlete

⋆ The velocity of the athlete

${\bigstar \:{\pmb{\sf{\underline{Solution...}}}}}$

⋆ The average speed of athlete = 20 m/s

⋆ The velocity of the athlete = 0 m/s

${\bigstar \:{\pmb{\sf{\underline{Using \: concepts...}}}}}$

⋆ Formula to find average speed

⋆ Formula to find velocity

${\bigstar \:{\pmb{\sf{\underline{Using \: formulas...}}}}}$

${\small{\underline{\boxed{\sf{\star \: Average \: speed \: = \dfrac{Distance}{Time}}}}}}$

${\small{\underline{\boxed{\sf{\star \: Velocity \: = \dfrac{Displacement}{Time}}}}}}$

${\bigstar \:{\pmb{\sf{\underline{Full \: Solution...}}}}}$

~ As it's given that He runs a distance of 100 m in 5 second "before returning to his original position" Henceforth, the displacement be 0 as because he returned to his original position.

~ Now let's find average speed by using suitable formula.

$:\implies \sf Average \: speed \: = \dfrac{Distance}{Time} \\ \\ :\implies \sf Average \: speed \: = \dfrac{100}{5} \\ \\ :\implies \sf Average \: speed \: = \cancel{\dfrac{100}{5}} \\ \\ :\implies \sf Average \: speed \: = 20 \: m/s$

~ Now let's find the velocity by using suitable formula.

$:\implies \sf Velocity \: = \dfrac{Displacement}{Time} \\ \\ :\implies \sf Velocity \: = \dfrac{0}{5} \\ \\ :\implies \sf Velocity \: = 0 \: m/s$

${\bigstar \:{\pmb{\sf{\underline{Explore \: more...}}}}}$

Distance = It is the length of actual path covered by a moving object in a given time interval.

Displacement = Shortest distance covered by a body in a definite direction is called displacement.

→ Displacement may be positive, negative or zero whereas distance is always positive.

→ Distance is a scaler quantity whereas displacement is a vector quantity both having the same unit.

→ In general magnitude of displacement ≤ Distance.

Speed = Distance travelled by a moving object in unit time interval is called speed i.e., speed Distance/Time. It's scaler quantity and it's SI unit is metre/second

Velocity = Velocity of moving object is defined as the displacement of the object in unit time interval i.e., velocity = Displacement/Time. It's vector quantity and it's SI unit is metre/second.

Acceleration = Acceleration of an object is defined as the rate of change of velocity of the object i.e., Acceleration = Change in velocity/Time. It's vector quantity and SI unit is m/s²