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(A) At $t = 1$,the positions of particle $A$ are:$x_A(1) = 3(1) = 3$$y_A(1) = 1$So,the position of particle $A$ is $(3, 1)$.At $t = 1$,the positions o

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$5$

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(A) At $t = 1$,the positions of particle $A$ are:$x_A(1) = 3(1) = 3$$y_A(1) = 1$So,the position of particle $A$ is $(3, 1)$.At $t = 1$,the positions of particle $B$ are:$x_B(1) = 6$$y_B(1) = 2 + 3(1)^2 = 2 + 3 = 5$So,the position of particle $B$ is $(6, 5)$.The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Substituting the values:$d = \sqrt{(6 - 3)^2 + (5 - 1)^2}$$d = \sqrt{3^2 + 4^2}$$d = \sqrt{9 + 16} = \sqrt{25} = 5$.

Two particles $A$ and $B$ are moving in the $XY$-plane. Their positions vary with time $t$ according to the relations:
$x_A(t) = 3t, \quad x_B(t) = 6$
$y_A(t) = t, \quad y_B(t) = 2 + 3t^2$
What is the distance between the two particles at $t = 1$?

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Two particles $A$ and $B$ are moving in the $XY$-plane. Their positions vary with time $t$ according to the relations:$x_A(t) = 3t, \quad x_B(t) = 6$$y_A(t) = t, \quad y_B(t) = 2 + 3t^2$What is the distance between the two particles at $t = 1$?