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Q.

Two particles 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+3t²
The distance between two particles at t = 1 is

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a

5

b

3

c

4

d

$\sqrt{12}$

answer is A.

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Detailed Solution

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At t = 1, x_A = 3. x_B = 6, Y_A = 1 and y_B = 5
So, distance= $\sqrt{{(x_{B} - x_{A})}^{2} + {(y_{B} - y_{A})}^{2}}$
$= \sqrt{3^{2} + 4^{2}} = 5$

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Two particles A and B are moving in XY-plane. Their positions vary with time t according to relationxA (t)=3t, XB (t)=6 YA (t)=t, YB(t)=2+3t2The distance between two particles at t = 1 is