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A cyclist is riding with a speed of `27 km h^(-1)`. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate `0.5 ms^(-1)`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?A. `0.86 m//s^(2)`B. `0.43 m//s^(2)`C. `1.24 m//s^(2)`D. `1.76 m//s^(2)` |
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Answer» Correct Answer - A `v=27xx(5)/(18)=(45)/(6)=(15)/(2)` ` a=sqrt(a^(2)+((v^(2))/(r))^(2))=sqrt(0.25+(15xx15)/(4xx80))` `=sqrt(0.25+(45)/(64))` `=sqrt(0.25+0.49)=sqrt(0.25+0.70)=sqrt(0.95)` `0.86m//s^(2)` |
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