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ICSE Class 10 Answered

A circular track has a circumference of 314 m with AB as diameter.A cyclist travels from A to B along circular path with a velocity of constant magnitude of 15.7 m/s.Find:(1) distance moved by the cyclist (2) displacementof the cyclist.(3)average velocity of the cyclist.(4)average acceleration of the cyclist?
Asked by nisha_vini29 | 03 Jul, 2017, 12:19: PM
answered-by-expert Expert Answer
begin mathsize 12px style Circumference space equals space 2 πR space equals space 341 space straight m rightwards arrow On space using space the space above space we space get space straight R space equals space 50 space straight m Given space that space straight v space equals space 15.7 space straight m divided by straight s left parenthesis straight a right parenthesis space Distance space moved space equals space to space πR space equals space 3.14 space cross times space 50 space equals space 157 space straight m left parenthesis straight b right parenthesis space Displacement space equals space Diameter space equals space 2 space straight R space equals space 100 space straight m left parenthesis straight c right parenthesis space Time space taken space by space the space cyclist space to space reach space form space one space end space of space the space diameter space to space the space other space end space equals space fraction numerator 157 over denominator 15.7 end fraction space equals space 10 space straight s rightwards arrow We space use space the space equation colon space Average space velocity space equals space Displacement over Time space equals space 100 over 10 space equals space 10 space straight m divided by straight s left parenthesis straight d right parenthesis space Average space acceleration space equals space fraction numerator Change space in space velocity over denominator Time space end fraction space equals space fraction numerator straight v subscript 2 space minus space straight v subscript 1 over denominator straight t end fraction rightwards arrow space Average space acceleration space equals space fraction numerator 15.7 space minus space left parenthesis negative 15.7 right parenthesis over denominator 10 end fraction rightwards arrow space Average space acceleration space equals space 3.14 space straight m divided by straight s squared end style
Answered by Yashvanti Jain | 03 Jul, 2017, 03:24: PM
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ICSE 10 - Physics
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