Matter In Our Surroundings

Question

A circular cycle track has a circumference of 314 m with AB as one of its diameter. A cyclist travels from A to B along the circular path with a velocity of constant magnitude 15.7 m/s. Find:
(a) the distance moved by the cyclist.
(b) the displacement of the cyclist if AB represents north-south direction.
(c) the average velocity of the cyclist.

Solution

Given, a circular track of circumference 314 m.

$G i v e n comma space space C i r c u m f e r e n c e comma space 2 pi r space equals space 314 space straight m therefore space space space space space space R a i d u s comma space straight r space equals fraction numerator 314 over denominator 2 straight pi end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 314 over denominator 2 cross times 3.14 end fraction space equals space 50 space straight m. left parenthesis straight a right parenthesis space D i s t a n c e space m o v e d comma space straight s space equals space 1 half cross times space C i r c u m f e r e n c e space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half cross times 314 space equals space 157 space straight m. left parenthesis straight b right parenthesis space D i s p l a c e m e n t space equals space stack A B with rightwards arrow on top space equals space 2 straight r space equals space 100 space straight m D i s p l a c e m e n t space i s space i n space n o r t h minus s o u t h space d i r e c t i o n. space left parenthesis straight c right parenthesis space T i m e space t a k e n space i n space g o i n g space f r o m space straight A space t o space straight B space equals space fraction numerator 157 space straight m over denominator 15.7 space straight m divided by straight s end fraction space equals space 10 space straight s$
$therefore space space A v e r a g e space v e l o c i t y space space equals space fraction numerator D i s p l a c e m e n t over denominator T i m e end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 100 over 10 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 10 space straight m comma space space i n space n o r t h minus s o u t h space d i r e c t i o n. space$

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