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Two particle A and B are moving in XY-plane. Their positions vary with time t according to relation `x_(A)(t)=3t, x_(B)(t)=6` `y_(A)(t)=t, y_(B)(t)=2+3t^(2)` The distance between two particle at `t=1` is :A. 5B. 3C. 4D. `sqrt(12)` |
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Answer» Correct Answer - 1 At `t=1, x_A = 3, x_B = 6, y_A = 1 and y_B = 5` so distance `= sqrt((x_B- x_A)^(2) + (y_B-y_A)^(2))` `= sqrt((6-3)^(2) + (5-1)^(2))= sqrt(3^(2)+4^(2))= 5` |
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