InterviewSolution
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InterviewSolution
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A car accelerates from rest at a constant rate `alpha` for some time, after which it decelerates at a constant rate `beta,` to come to rest. If the total time elapsed is t seconds. Then evalute (a) the maximum velocity reached and (b) the total distance travelled.A. `(alpha beta t^(2))/(4(alpha +beta))`B. ` (alpha beta t^(2))/(2(alpha +beta))`C. `(alpha beta t^(2))/(alpha +beta)`D. `(4alpha beta t^(2))/(alpha +beta)` |
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Answer» Correct Answer - B From `A` to `B` applying `sut +(1)/(2) at^(2)`, we get `S_(1) =0 xx t_(1) +(1)/(2) alph t_(1)^(2) =(1)/(2) (alpha t_(1)) t_(1) =(1)/(2) v_(0)t_(1)` (1) Sinmilarly from `B` to `C`: `S_(2) =v_(0) t_(2) +(1)/(2) (-beta) t_(2)^(2) =v_(0)t_(2) -(1)/(2) (betat_(2)) t_(2) =(1)/(2) v_(0) t_(2) (2) Now total distance travelled is `S_(1) +S_(2) =(1)/(2) v_(0)t_(1) +(1)/(2) v_(0)t_(2) =(1)/(2) v_(0) (t_(1) _t_(2))` `=(1)/(2) (alph beta t)/(2 alpha+beta) t =(alpha beta t^(2))/(2 (alpha+beta) t =(alpha beta t^(2))/(2(alpha +beta)` Also, we can find (i) and (ii) by using the formula, `s=(u+v)/(2) t`. |
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