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A particle moves in a straight line so that `s=sqrt(t)`, then its acceleration is proportional toA. `("velocity")^3`B. velocityC. `("velocity")^2`D. `("velocity")^(3//2)` |
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Answer» Correct Answer - A Given, `s=sqrt(t)` On differenting w.r.t., we get `(ds)/(dt)=(1)/(2sqrt(1))rArrv=(1)/(2sqrt(t))` Again differentiating. We get `(dv)/(dt)=-(1)/(2.2t^(3//2))` `rArr a=-(2)/((2sqrt(t))^3)` `rArr a=-2v^3rArr a prop v^3` . |
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