Question

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

Answer 1

Answer:

The distance covered by the athlete is 2200 m and the displacement at the end of 2 min 20s is 200 m.

Explanation:

Given : Diameter of circular track = 200 m

Radius 'r' = 200/2 = 100 m

Time taken by the athlete to complete one round of circular track = 40 seconds

Now, time = 2 min 20s = (2 × 60) + 20 = 140 seconds (Since, 1 minute = 60 seconds)

Number of rounds covered by the athlete = Total time / Time taken to complete one round

= 140/40 = 3.5Circumference of a circle = 2πr

Distance covered in 1 round = Circumference of the circular track

Thus, the circumference of the circular track = 2 × 22/7 × 100

Distance traveled in 140 seconds = Number of rounds covered × Circumference of the track

= 3.5 × 2 × 22/7 × 100

= 0.5 × 2 × 22 × 100

1. Thus, the distance covered by the athlete is = 2200 mNow, to calculate the displacement let's observe the following diagram.
2. Let the athlete start from point A and reach point B after 140 seconds.
4. Thus, the displacement will be equal to the length of the line segment AB.
6. Length of AB = Diameter of the circle = 200 m
8. Thus, the displacement of the athlete is 200 m
10. Therefore, the distance covered by the athlete to complete a circular track of a diameter 200 m at the end of 2 min 20 seconds is 2200 m and the displacement is 200 m.

Answer 2

Given :

• Diameter of circular track = 200 .
• Radius of circular track = $\sf \cancel{ \dfrac{200}{2}}$ = 100 m
• Time taken by Athlete to complete one round = 40 s

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To find :

• Distance covered and displacement at the end of 2 min 20 s.

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Solution :

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★ 2 min 20 s = 140 seconds

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• In 40 sec athlete complete one round

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So, In 140 sec he will complete = $\sf \cancel{ \dfrac{140}{4}} = 3.5\;rounds\;\;\;\;\:\:\:\bigg\lgroup\bf 2\;min\;20\;s = 140\;sec\bigg\rgroup$

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• One round = Circumference of Circle

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Therefore,

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• Distance covered in 140 sec = $\sf 2 \pi r \times 3.5$

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$:\implies\sf 2 \times 3.14 \times 100 \times 3.5$

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$:\implies\sf 2 \times \dfrac{314}{ \cancel{100}} \times \cancel{100} \times 3.5$

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$:\implies\sf 2 \times 314 \times 3.5$

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$:\implies{\underline{\boxed{\bf{\pink{2200\;m}}}}}\;\bigstar$

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$\therefore$ Hence, the distance covered at the end of 2 min 20 s is 2200 m.

And Displacement is 200 m (diameter) as it return to initial position again.