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Displacement-time equation of a particle moving along x-axis is `x=20+t^3-12t` (SI units) (a) Find, position and velocity of particle at time t=0. (b) State whether the motion is uniformly accelerated or not. (c) Find position of particle when velocity of particle is zero. |
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Answer» Correct Answer - A (a) `x=20+t^3-12t` …(i) At t=0, `x=20+0-0=20m` Velocity of particle at time t can be obtained by differentiating Eq. (i) w.r.t. time i.e. `v=dx/dt=3t^2-12` …(ii) At t=0, `v=0-12 =-12m//s` (b) Differentiating Eq. (ii) w.r.t. time t, we get the acceleration `a=(dv)/(dt) =6t` As acceleration is a function of time, the motion is non-uniformly accelerated. (c) Substituting v=0 in Eq. (ii), we have `0=3t^2-12` Positive value of t comes out to be 2s from this equation. Substituting `t=2s` in Eq. (i), we have `x=20+(2)^3-12(2)` or `x=4m` |
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