InterviewSolution
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InterviewSolution
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The velocity of a particle moving in the `x` direction varies as `V = alpha sqrt(x)` where `alpha` is a constant. Assuming that at the moment `t = 0` the particle was located at the point `x = 0`. Find the acceleration. |
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Answer» `v=alphasqrt(x)=alphax^(1//2) [Here v=fn(x)]` `(dv)/(dx)=alpha.(1)/2x^(-1//2)` Acceleration `a=v(dv)/(dx)=alphax^(1//2) alpha/2x^(-1//2)=alpha^(2)/2` Acceleration is constant. Using equation of motion, `v=u+at=0+alpha^(2)/2t` `=(alpha^(2)t)/2` |
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