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

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Chetan Santha

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Ans: Time taken = 2 min 20 sec = 140 sec. Radius, r = 100 m. InTime to complete one round = 40 sec So, Round completed in 140 sec = 140 ÷ 40 = 3.5 (three and a half) round. Or, distance covered = 2 x 22/7 x 100 x 3.5 = 2200 m. At the end, the athlete cavers half round and will be at a distance...
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Ans: Time taken = 2 min 20 sec = 140 sec. Radius, r = 100 m. InTime to complete one round = 40 sec So, Round completed in 140 sec = 140 ÷ 40 = 3.5 (three and a half) round. Or, distance covered = 2 x 22/7 x 100 x 3.5 = 2200 m. At the end, the athlete cavers half round and will be at a distance = diameter of circle. ie displacement = 200m. read less
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Science Tutor

200m is the answer
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Classes for IX -X Mathematics & Science

Distance covered = 2198m, Displacement = 200m
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one rround circular tract=2*pi rad angular speed= w= 2xpi/40=0.157 rad/sec dispalcement , theta=w*t= 21.98 radians distance covered=140/40 X(2XpiX100)=2200m which is coveered in 40 sec distance covered in 2 min 20 sec=140sec =14/40*1257.14
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here radius r=100m 2min 20sec=140sec. no of complete rotation=140/40=3.5=3+1/2. so he will end up by the time 140 sec just opposite along the diagonal, so displacement = length of diameter=200 m. total distance covered=3*(2*pi*r)+pi*r=7*pi*r=2199.1 m
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HI For a circular track : velocity(v)= circumference/time v=3.14*200/40=15.7m/s distance covered in 2min 20 sec(140sec)= velocity * time = 140*15.7=2198m 40 seconds are required to complete 1 rotation therefore in 140 seconds, rotations completed will be =140*1/40=3.5 Thus this means three complete...
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HI For a circular track : velocity(v)= circumference/time v=3.14*200/40=15.7m/s distance covered in 2min 20 sec(140sec)= velocity * time = 140*15.7=2198m 40 seconds are required to complete 1 rotation therefore in 140 seconds, rotations completed will be =140*1/40=3.5 Thus this means three complete rotations and one half rotation. Therefore athlete will be diametrically opposite to his position and his displacement will be equal to length of diameter i.e. = 200m read less
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Here r=200/2 m=100 m. 2 min 20 sec= 140 sec. No. Of complete rotation 140/40= 3.5= 3+1/2. From definition of displacement it will be diameter of the circle i.e 200 m. Total distance covered= +?r=7?r=2221.8 m.
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In the given time athlete will complete 3.5 round so disp will be 200m and dist 2198m
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IITian Tutor

Displacement will be 200m, while distance = 3*2*pi*100 + pi*100 = 2198m
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M.Sc. Biomedical Genetics, B.Sc. Biotechnology

Here we have, Diameter = 200 m, therefore, radius = 200m/2 = 100 m Time of one rotation = 40s Time after 2m20s = 2 x 60s + 20s = 140s Distance after 140 s = ? Displacement after 140s =? Motion Ex1 Motion Ex2 (a) Distance after 140s We know that,distance=velocity ×time ?distance=15.7m/s...
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Here we have, Diameter = 200 m, therefore, radius = 200m/2 = 100 m Time of one rotation = 40s Time after 2m20s = 2 x 60s + 20s = 140s Distance after 140 s = ? Displacement after 140s =? Motion Ex1 Motion Ex2 (a) Distance after 140s We know that,distance=velocity ×time ?distance=15.7m/s ×140 s = 2198 m (b) Displacement after 2 m 20 s i.e. in 140 s Since,rotatin in 40 s=1 Motion Ex3 Therefore, in 3.5 rotations athlete will be just at the opposite side of the circular track, i.e. at a distance equal to the diameter of the circular track which is equal to 200 m. Therefore, Distance covered in 2 m 20 s = 2198 m And, displacement after 2 m 20 s = 200m 2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C? Answer:- Motion Ex4 Here we have, Distance from point A to B = 300 m Time taken = 2 minute 30 second = 2 x 60 + 30 s = 150 s Distance from point B to C = 100 m Time taken = 1 minute = 60 s (a) Average speed and velocity from point A to B Motion Ex5 (b) Average speed and velocity from B to C Motion Ex6 Therefore,average velocity=1.66 m/s west read less
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