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Find the equation of trajectory of a particle whose velocity components are `v_(x)=2x+1, v_(y)=2y+3` Given that particle starts from rest from origin. |
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Answer» `v_(x)=2x+1`, `(dx)/(dt)=2x+1` `int_(0)^(x) (dx)/(2x+1)=int_(0)^(t)dt` `(1)/(2)ln(2x+1)=t " "` ....(1) `v_(y)=2y+3` `int _(0)^(y)(dy)/(2y+3)=int_(0)^(t)dt` `(1)/(2) "ln "((2y+3))/(3)=t" "` ...(2) from (1) and (2) `(1)/(2) ln (2x+1)=(1)/(2) "ln" ((2y+3))/(3)` `2x+1=((2y+3))/(3)` `y=3x` |
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