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The motion of a particle along a straight line is described by the function `x=(2t -3)^2,` where x is in metres and t is in seconds. Find (a) the position, velocity and acceleration at `t=2 s.` (b) the velocity of the particle at origin. |
Answer» (i) Posiotion of the particle, `x = (2t-3)^(2)` At `t=2s, x=(2xx2-3)^(2)=1.0 m` Velocity, `v=(dx)/(dt)=4(2t-3)ms^(-1)` At `t = 2 s, v= 4(2xx2-3)4ms^(-1)` and acceleration of the particle, `a=(dv)/(dt)=8ms^(-2)` (ii) At origin, x = 0 or `(2t-3)^(2)=0` Velocity of the particle at origin, `v=4xx0=0` |
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