InterviewSolution
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InterviewSolution
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The position of a particle moving along x-axis is related to time `t` as follow: `x=2 t^(2)-t^(3)`, where `x` is in meters and `t` is in seconds. a. What is the maximum positive displacement of the particle along the `x` axis and at what instant does it attain it? b. Describe the motion of the particle. c. What is the destamce covered in the first three seconds? d. What is its desplacement in the first four seconds ? e. What is the particle`s average speed and average veloctry in the first `3` seconds ? f. What particles instantaneous acceleration at the instant of its maximum positiv `x` displacement? ltbtgt g. What is the average acceleration between the interval `t=2` to `t=4 s`?. |
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Answer» The derivative of `x` w.r.t. time will give velocity of the particle in relation with time. `v=(dx)/(dt)=4t-3t^(2)` ..(i) The double derivative of `x` w.r.t. time will give velocity of the particle in relation with time. `a=(d^(2)x)/(dt^(2))=4-6t`.........(ii) a. For minimum and maximum displacement, `dx//dt=0`. Thus, `(dx)/(dt)=4t-3t^(2)=0` `t(4-3t)=0 rArr` either `t=0` or `t=(4)/(3) s` Also, for maxima, double derivativ `x` should be negative, i.e.,`(d^(2)x)/(dt^(2)) lt0`. `(d^(2)x)/(dt^(2))=4-6t` At`t=0: (d^(2)x)/(dt^(2))=4-(0)=4`(positive) And at t=(4)/(3) s: (d^(2)x)/(dt^(2))=4-6((4)/(3))=4-8=-4`(negative) (negative) Threfore, the porticle is at maximum displacement at `t=(4)/(3) s` and the corresponding displacement is `x_(max)=2((4)/(3))^(2)-((4)/(3))^(3)=(32)/(27)m`. |
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