Two particles A and B are moving in X Y-plane | vedclass

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Question

(a)

At $t=1, x_A=3, x_B=6, y_A=1$ and $y_B=5$

so distance $=\sqrt{\left( x _{ B }- x _{ A }\right)^2+\left( y _{ B }- y _{ A }\right)^2}$ $=\sqr

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(a)

At $t=1, x_A=3, x_B=6, y_A=1$ and $y_B=5$

so distance $=\sqrt{\left( x _{ B }- x _{ A }\right)^2+\left( y _{ B }- y _{ A }\right)^2}$ $=\sqrt{(6-3)^2+(5-1)^2}=\sqrt{3^2+4^2}=5$

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 :

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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 :