Two particles A and B are moving in the XY pl | vedclass

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(C) For particle $A$,the path is given by $y = x$.Differentiating with respect to time $t$,we get $\frac{dy}{dt} = \frac{dx}{dt}$,which implies $v_{y_

Vedclass · Accepted Answer

$3\sqrt{2} \ m/s$

Vedclass · Answer

(C) For particle $A$,the path is given by $y = x$.Differentiating with respect to time $t$,we get $\frac{dy}{dt} = \frac{dx}{dt}$,which implies $v_{y_A} = v_{x_A}$.Since particle $B$ moves along the $X$-axis with a uniform speed of $3 \ m/s$,its velocity is $\vec{v}_B = 3 \hat{i}$.Given that the $X$-coordinates of $A$ and $B$ are always equal,the $X$-component of the velocity of $A$ must be equal to the velocity of $B$,so $v_{x_A} = 3 \ m/s$.Since $v_{y_A} = v_{x_A}$,we have $v_{y_A} = 3 \ m/s$.Thus,the velocity vector of particle $A$ is $\vec{v}_A = 3 \hat{i} + 3 \hat{j}$.The speed of particle $A$ is the magnitude of its velocity vector: $|\vec{v}_A| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \ m/s$.

Column $I$	Column $II$
$(A)$ Displacement	$(p)$ $8 \sin 2$
$(B)$ Distance	$(q)$ $4$
$(C)$ Average velocity	$(r)$ $2 \sin 2$
$(D)$ Average acceleration	$(s)$ $4 \sin 2$

Two particles $A$ and $B$ are moving in the $XY$ plane. Particle $A$ moves along a line with equation $y = x$,while particle $B$ moves along the $X$-axis such that their $X$-coordinates are always equal. If particle $B$ moves with a uniform speed of $3 \ m/s$,what is the speed of particle $A$?

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