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A particle moves in xy plane with a velocity by `vec(v)=(8t-2)hat(i)+2hat(j)`. If it passes through the point (14, 4) at t = 2 sec, then give equation of the path.A. `x + y - y^(2) = 2`B. `y^(2)+x=4`C. `x-y-y^(2)=2`D. `y + x + y^(2)=2` |
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Answer» Correct Answer - A `V = 8t -2` `intdx = int(8t-2)dt` `x = 4t^(2)-2t +c` At `t =2,x = 14 implies " "14=4(4) -2(2)+c " implies "c = 2` `x = 4t^(2)-2t+2,v_(y)=2` `intdy int2dt` `y = 2t + c` At `t =2, y = 4 " implies "c = 0` ` y = 2t` Equation `implies " " x = (2t)^(2)-2t +2` `implies" " x=y^(2)-y+2` `implies " "x + y-y^(2) =2` |
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