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Two particles $A$ and $B$ are moving in $X Y$-plane.
Their positions vary with time $t$ according to relation :
$x_A(t)=3 t, \quad x_B(t)=6$
$y_A(t)=t, \quad y_B(t)=2+3 t^2$
Distance between two particles at $t =1$ is :
A
$5$
B
$3$
C
$4$
D
$\sqrt{12}$
Solution
(a)
At $t=1, x_A=3, x_B=6, y_A=1$ and $y_B=5$
so distance $=\sqrt{\left( x _{ B }- x _{ A }\right)^2+\left( y _{ B }- y _{ A }\right)^2}$ $=\sqrt{(6-3)^2+(5-1)^2}=\sqrt{3^2+4^2}=5$
Standard 11
Physics
