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
end of 2 minutes 20s ?​

Answer 1

Given :

• An athlete completes one round of a circular track of diameter 200 metre in 40 seconds.

To Find :

• The distance covered and the displacement at the end of 2 minutes 20 second.

Solution :

Here,

Total time = 2 minutes 20 seconds

→ 2 × 60 seconds + 20 seconds

→ 120 seconds + 20 seconds

→ 140 seconds

Now,

In 40 s, the no. of rounds completed = 1

So,

In 140 s,the no. of rounds completed = $\sf\dfrac{1}{40}$ × 140

In 140 s, the no. of rounds completed = 3.5

• Distance covered in 3.5 rounds :

When diameter of the track is 200 m, radius will be 100 m

Now,

[Distance covered in 1 round = Circumference of the circle]

→ 2 × 22/7 × 100

→ 4400/7

→ 628.57

Total Distance covered = 628.57 × 3.5 m

→ 2200

$\therefore$ Distance covered at the end of 2 minutes 20 seconds is 2200 m.

• Displacement covered in 3.5 rounds :

The athlete makes three and a half rounds now If the athlete starts from point A, then after completion of 3 rounds he'll be at the point A again. And when again start from point A and makes the remaining half round, he'll reach point B. So, the displacement of athlete will be equal to diameter of the circular track (Point B is diametrically opposite to point A)

$\therefore$ Displacement covered at the end of 2 minutes 20 seconds is 200 m.

• Refer to the attachment!

Answer 2

$\huge {\underline {\mathfrak { Question }}}$

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 20s ?

$\huge {\underline {\mathfrak { Given }}}$

• He completes one round of a circular track of diameter 200m in 40s.

$\huge {\underline {\mathfrak { To \: find }}}$

• Distance covered at the end of 2minutes 20 seconds.

• Displacement at the end of 2minutes 20 seconds.

$\huge {\underline {\mathfrak { Calculation}}}$

$☞$ Diameter of the circular track = 200m

☞ So radius = $\sf\frac{\cancel {200}}{\cancel {2}}= 100m$

☞Time taken to complete one round = 40s

☞Total time given = 2 minutes 30 seconds

$\: \: \: \: \: \: \: \:$= ( 2 × 60 + 30s )

$\: \: \: \: \: \: \: \:$= ( 120s + 30s )

$\: \: \: \: \: \: \: \:$= ( 140s )

In 40 seconds, athelete completed 1 round.

☞40 s = 1 round

☞1s = $\sf { \frac { 1}{40} round }$

☞140s = $\sf { \frac { 1}{ \cancel {4}^{2} \cancel {0}} \times \cancel { 14}^{ 7} \cancel {0} }$

$\sf { \frac {1}{ \cancel 7} \times \cancel{14}^{3.5} = 3.5 \: rounds }$

Therefore, in 2min 20s,the athlete completed 3.5rounds.

Now,

☞Distance covered by the athelete in one round = 2πr ( circumference )

= $\sf {2 \times \frac{22}{7} \times 100}$

☞Therefore, distance covered in 3.5 rounds=

$\implies \sf\green {3.5 \times 2 \times \frac{22}{7} \times 100}$

$\implies \sf { \frac{ { \cancel{35}}^{5} }{1 \cancel0} \times 2 \times \frac{22}{ \cancel7} \times 10 \cancel0}$

$\implies \sf {5 \times 2 \times 22 \times 100}$

$\implies\sf {10 \times 22 \times 100}$

$\implies \sf\red {220 \times 100}$

$\implies \sf\red {2200m}$

Total distance covered in 3.5 round is 2200 m.

_____..

☞ Displacement = AB = 200m

[ Here, it is 200m because the athelete completes 3.5 rounds, simply we can say that he completes 3 rounds and then he comes to again on his position where he was at starting and then he completes half of the round,that means means its final velocity (v) is 200m(diameter) and initial velocity(u) was 0 since it was at rest,so according to formula of displacement-

v-u = displacement

200m - 0m = 200m ]

$\huge {\underline {\mathfrak { Required \: answers}}}$

• Distance covered at the end of 2minutes 20 seconds = 2200m

• Displacement at the end of 2minutes 20 seconds = 200m