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A car is moving along a stight horizontal road with a speed `v_(0)` . If the coefficient of friction between the tyres and the road is mu, the shortest distance in which the car can be stopped is.A. `(v_(0)^(2))/(2mug)`B. `(v_(0))/(mug)`C. `((v_(0))/(mug))^(2)`D. `(v_(0))/(mu)` |
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Answer» Correct Answer - A Retarding force `F=ma=muR=mumg` :. `a=mug` Now from equation of motion `v^(2)=u^(2)-2as` implies `0=u^(2)-2as` implies `s=(u^(2))/(2a)=(u^(2))/(2mug)` implies `s=(v_(0)^(2))/(2mug)` |
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