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Two particle A and B are moving in XY-pl...

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

5

B

3

C

4

D

`sqrt(12)`

Text Solution

Verified by Experts

The correct Answer is:
A
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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