A GIRL CYCLES AROUND A RECTANGULAR PATH OF LENGTH 20 m and breadth 10 m in 60 sec . CALCULATE THE DISTANCE AND DISPLACEMENT COVERED BY THE GIRL AFTER 3 MIN 30 SEC ?
princey:
DOES ANYONE KNOW HOW TO DO IT?
distance is 210m and displacement is 10*(5)^1/2.remember distance is not exactly the displacement. the latter is the shortest distance between 2 points.here the displacement is hypotenuse.
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Answered by
0
SPEED=DISTANCE/TIME=perimeter of he rectangle/time=2(10+20)/60=60/60=1 ms-1
so after 3 min 30 sec (3(180)+30=540+30=570 s) distance moved
=570(1)=570 m
Therefore displacement produced is 570-540=30 m
(as it has completed four cycles then again covered 30)
(so gap between initial and final is 30)
so after 3 min 30 sec (3(180)+30=540+30=570 s) distance moved
=570(1)=570 m
Therefore displacement produced is 570-540=30 m
(as it has completed four cycles then again covered 30)
(so gap between initial and final is 30)
Answered by
2
for 1 minute she completes whole rectangle hence displacement after 1 minute(60 seconds ) is 0 after 3 minutes displacement is 0 for 30 seconds she covers half the rectangle therefore displacement is the length of diagonal of rectangle
which is
=
(by pythagorus theroem of right angled triangles
distance is length of path travelled
which is 3*(2*(20+10))+20+10=180+30=210m
which is
distance is length of path travelled
which is 3*(2*(20+10))+20+10=180+30=210m
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