Q. 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?
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Q. 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?
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The diameter of the circular track is given as 200 m. That is, 2r=200 m.
The diameter of the circular track is given as 200 m. That is, 2r=200 m.From this radius is calculated as Radius, r = 100 m.
The diameter of the circular track is given as 200 m. That is, 2r=200 m.From this radius is calculated as Radius, r = 100 m.In 40 sec the athlete complete one round. So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.
The diameter of the circular track is given as 200 m. That is, 2r=200 m.From this radius is calculated as Radius, r = 100 m.In 40 sec the athlete complete one round. So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.One round is considered as the circumference of the circular track.
The diameter of the circular track is given as 200 m. That is, 2r=200 m.From this radius is calculated as Radius, r = 100 m.In 40 sec the athlete complete one round. So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.One round is considered as the circumference of the circular track.The distance covered in 140 sec = 2πr×3.5=2×3.14×100×3.5=2200 m.
The diameter of the circular track is given as 200 m. That is, 2r=200 m.From this radius is calculated as Radius, r = 100 m.In 40 sec the athlete complete one round. So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.One round is considered as the circumference of the circular track.The distance covered in 140 sec = 2πr×3.5=2×3.14×100×3.5=2200 m.For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.
The diameter of the circular track is given as 200 m. That is, 2r=200 m.From this radius is calculated as Radius, r = 100 m.In 40 sec the athlete complete one round. So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.One round is considered as the circumference of the circular track.The distance covered in 140 sec = 2πr×3.5=2×3.14×100×3.5=2200 m.For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.At the end of his motion, the athlete will be in the diametrically opposite position. That is, displacement = diameter = 200 m.
The diameter of the circular track is given as 200 m. That is, 2r=200 m.From this radius is calculated as Radius, r = 100 m.In 40 sec the athlete complete one round. So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.One round is considered as the circumference of the circular track.The distance covered in 140 sec = 2πr×3.5=2×3.14×100×3.5=2200 m.For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.At the end of his motion, the athlete will be in the diametrically opposite position. That is, displacement = diameter = 200 m.Hence, the distance covered is 2200 m and the displacement is 200 m.
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The diameter of the circular track is given as 200 m. That is, 2r=200 m.
From this radius is calculated as Radius, r = 100 m.
In 40 sec the athlete complete one round. So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.
One round is considered as the circumference of the circular track.
The distance covered in 140 sec = 2πr×3.5=2×3.14×100×3.5=2200 m.
For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.
At the end of his motion, the athlete will be in the diametrically opposite position. That is, displacement = diameter = 200 m.
Hence, the distance covered is 2200 m and the displacement is 200 m.