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A particle move a distance `x` in time `t` according to equation `x = (t + 5)^-1`. The acceleration of particle is alphaortional to.A. `("velocity")^(3//2)`B. `("distance")^(2)`C. `("distance")^(-2)`D. `("velocity")^(2//3)` |
Answer» Correct Answer - A Given, `x=(1)/(t+5)` Differentiating both sides w.r.t. t, we get `(dx)/(dt)=-(1)/((t-5)^(2))xx1` `therefore` Velocity, `upsilon =(dx)/(dt)=-(1)/((t+5)^(2))` ….(i) Again differentiating both sides w.r.t. t, we get `(d upsilon)/(dt)=(1xx2)/((t+5)^(3))(1)=(2)/((t+5)^(3))` ....(ii) Now, from Eqs. (i) and (ii), we get `(d upsilon)/(dt)=-2upsilonxx(1)/((t+5))` `=(-2upsilon)(sqrt(upsilon))=-2(upsilon)^(3//2)` `therefore` Acceleration, `a=(d upsilon)/(dt)=-2upsilon^(3//2)rArr a prop ("velocity")^(3//2)` |
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