InterviewSolution
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InterviewSolution
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A horizontal force `F` of variable magnitude and constant direction acts on a body of mass `m` which is initially at rest at a point `O` on a smooth horizontal surface. The magnitude of `F` is given by `F=beta+alphat` where `t` is the time for which the force has been acting on the distance of the body from `O` at time `t` , then `s` is equal to.A. `(1)/(2m)(betat+alphat^(2))t`B. `(1)/(2m)(beta+alphat^(2))`C. `((beta+alphat)t^(2))/(2m)`D. `t^(2)/(6m)(3beta+alphat)` |
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Answer» Correct Answer - D `a=(dv)/(dt)=(F)/(m)=(beta+alphat)/(m)` `int_(0)^(v)dv=int_(0)^(t)((beta+alphat)/(m))dt` implies `v=(betat)/(m)+(alphat2)/(2m)` Now `intds=intvdt` implies `int_(0)^(s)ds=int_(0)^(t)((betat)/(m)+(alphat^(2))/(2m))dt` implies `s=(t^(2))/(6m)(3beta+alphat)` |
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