Question

an athlete completes one round of a circular track of diameter 200m in 40 seconds what will be the distance covered and the displacement at the end of 2 minutes 20 seconds. please answer me fast!!!​

Answer 1

GIVEN :

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

i.e. diameter, d = 200 m.

time taken, t = 40 sec.

TO FIND :

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

SOLUTION :

Here,

Am athlete completes one round in 40 sec.

In 2 mins 20 secs, mean 140 secs.

In 140 secs, the athlete will complete = 140/40 = 3.5 mean 3 and a half rounds.

Now,

we know that, the round in circular,

so, One round is considered as the circumference of the circular track.

• Circumference of the circular track is = 2 πr.

The distance covered in 140 secs = 2 πr × 3.5

= 2 × 3.14 × 100 × 3.5

= 2200 m.

•°• Displacement will be zero for the each round.

So, the displacement for three rounds will also be zero.

•°• The distance covered at the end of 2 minutes 20 seconds is 2200 m.

Answer 2

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 ?

Solution:

${\underline {\underline {\boxed {\rm {\orange { Given:}}}}}}$

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

${\underline {\underline {\boxed {\rm {\orange { To \: find:}}}}}}$

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

• Displacement at the end of 2minutes 20 seconds.

${\underline {\underline {\boxed {\rm {\orange { Calculation:}}}}}}$

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

$\sf { \frac {1}{ \cancel 2} \times \cancel{7}^{3.5} = 3.5 \: rounds }$

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 ]

${\underline {\underline {\boxed {\rm {\orange { Required \: answers:}}}}}}$

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

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

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