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A particle moves in x-y plane. The position vector of particle at any time t is vecr={(2t)hati+(2t^2)hatj} m. The rate of change of thetaat time t = 2 s (where theta is the angle which its velocity vector makes with positive x-axis) is |
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Answer» `2/17 "rad s"^(-1)` Comparing it with standard equation of position vector , `vecr=xhati+yhatj` , we get x=2t and `y=2t^2` `rArr v_x=(dx)/(dt)=2` and `v_y=(dy)/(dt)=4t` `THEREFORE tan THETA = v_y/v_x = (4t)/2=2t` Differentiating with respect to time we get , `(sec^2 theta ) (d theta)/(dt)=2` or `(1+tan^2 theta) (d theta)/(dt) = 2` or `(d theta)/(dt) =2/(1+4t^2)` at t=2 s , `((d theta)/(dt))=2/(1+4(2)^2)=2/17 "rad s"^(-1)` |
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