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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 We have, `s=sqrt(t)implies(ds)/(dt)=(1)/(2sqrt(t))and (d^(2)s)/(dt^(2))=(1)/(4t^(3//2))` `implies (d^(2)s)/(dt^(2))=-((2ds)/(dt))^(3)=-2((ds)/(dt))^(3)` Hence, Acceleration `prop` `("Velocity")^(3)`. |
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