Question

An athlete completes one round of a circular track of diameter 200m in 40s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Answer 1

Determine the number of rounds completed by the athlete in 2 minutes and 20 seconds. Determine the total distance travelled by multiplying the number of rounds completed by circumference of the circular track. Then find the displacement of the athlete from the number of rounds completed by the athlete.

Formula used:
The circumference $C$ of the circle in terms of its diameter is
$C = \pi d$ …… (1)
Here, $d$ is the diameter of the circle.

Complete step by step solution:
We have given that the diameter of the circular track on which the athlete moves is $200\,{\text{m}}$ and the time taken by the athlete to complete one round or time period of athlete is $40\,{\text{s}}$.
$d = 200\,{\text{m}}$
$T = 40\,{\text{s}}$
We have asked to determine the distance covered and displacement of the athlete in 2 minutes and 20 seconds.
Let us first determine the number of rounds of the circular track completed by the athlete in 2 minutes and 20 seconds.
Convert the unit of total time in the SI system of units.
$t = \left( {2\,\min } \right)\left( {\dfrac{{60\,{\text{s}}}}{{1\,{\text{min}}}}} \right) + \left( {\,20\,{\text{s}}} \right)$
$\Rightarrow t = \left( {120\,{\text{s}}} \right) + \left( {\,20\,{\text{s}}} \right)$
$\Rightarrow t = 140\,{\text{s}}$
Hence, the time 2 minutes and 20 seconds in the SI system of units is $140\,{\text{s}}$.
Divide time $t$ by time period $T$ to determine the number $n$ of the round completed.
$n = \dfrac{t}{T}$
Substitute $140\,{\text{s}}$ for $t$ and $40\,{\text{s}}$ for $T$ in the above equation.
$n = \dfrac{{140\,{\text{s}}}}{{40\,{\text{s}}}}$
$n = 3.5$
Hence, the number of rounds completed by the athlete in 2 minutes and 20 seconds is 3.5.
The distance travelled by the athlete in one round is equal to circumference of the circular track.
Hence, the distance travelled by the athlete in 3.5 rounds is
${\text{Distance}} = 3.5\pi d$
Substitute $3.14$ for $\pi$ and $200\,{\text{m}}$ for $d$ in the above equation.
${\text{Distance}} = 3.5\left( {3.14} \right)\left( {200\,{\text{m}}} \right)$
$\Rightarrow {\text{Distance}} = 2198\,{\text{m}}$
$\Rightarrow {\text{Distance}} \approx 2200\,{\text{m}}$
Hence, the distance covered by the athlete in the given time is $2200\,{\text{m}}$.
After each round, the athlete reaches to its original position making its displacement zero. Hence, the displacement of the athlete in the first three rounds is zero.
After the displacement of the athlete in the last half round, the athlete reaches to the diametrically opposite position compared to its original position. Hence, the displacement of the athlete in the last half round is approximately equal to the diameter of the circular track.
Hence, the displacement of the athlete in the given time is $200\,{\text{m}}$.
Thus, the distance covered and displacement of the athlete in the given time are and $200\,{\text{m}}$ respectively.

Note:
The actual distance travelled by the athlete is $2198\,{\text{m}}$ but we have rounded it to $2200\,{\text{m}}$. This rounding of the distance covered by the athlete makes the calculation of displacement of the athlete in the last half round of his travel easier and we can determine the displacement from the length of diameter of the circular track.