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A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as alpha=ksqrt(v),where k Is a positive constant.At the initial moment the velocity of the point is equal to V0.What distance will it take to cover that distance ? |
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Answer» Solution :Let `t_(0)` be the tine in which it comes to a stop. GIVEN that `-(dv)/(DT)=ksqrt(V) underset(0)overset(t_(0))INT kdt=underset(v_(0))overset(0)int-(dv)/(sqrt(V))` `therefore kt_(0)=2sqrt(v_(0)) ""therefore t_(0)=(2)/(k)sqrt(v_(0))` Now to find the distance covered before stopping. `(dv)/(dt)=(dv)/(DS)-(ds)/(dt)=v(dv)/(ds)` But, `(dv)/(dt)=-ksqrt(V)`, `thereforev(dv)/(ds)=-ksqrt(V)` `therefore sqrt(vdv)`=-kds `therefore underset(v_(0))overset(0)sqrt(vdv)=-underset(0)overset(s)int dsimpliess=(2)/(3k)V_(0)^((3)/(2))`. |
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