(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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(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$.

Class 11 Physics 3-2.Motion in Plane Motion In Two And Three Dimension

Medium

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 $t = 1$?

A
$5$
B
$3$
C
$4$
D
$\sqrt{12}$

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3-2.Motion in Plane

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Motion In Two And Three Dimension

$A$ particle moves in the $x-y$ plane with velocity $\vec{v} = a\hat{i} + bx\hat{j}$,where $a$ and $b$ are constants. If the particle was initially at the origin,find the equation of its trajectory.

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An ant is moving on a plane horizontal surface. The number of degrees of freedom of the ant will be .........

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Position of an ant ($S$ in metres) moving in $Y-Z$ plane is given by $S = 2t^2 \hat{j} + 5 \hat{k}$ (where $t$ is in seconds). The magnitude and direction of velocity of the ant at $t = 1 \ s$ will be:

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The position vector of a moving body at any instant of time is given as $\overrightarrow{r} = (5t^2 \hat{i} - 5t \hat{j}) \text{ m}$. The magnitude and direction of velocity at $t = 2 \text{ s}$ is,

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$A$ point moves in the $x-y$ plane according to $x = kt$ and $y = kt(1 - \alpha t)$,where $k$ and $\alpha$ are positive constants. The equation of the trajectory is:

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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 $t = 1$?

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$A$ particle moves in the $x-y$ plane with velocity $\vec{v} = a\hat{i} + bx\hat{j}$,where $a$ and $b$ are constants. If the particle was initially at the origin,find the equation of its trajectory.

An ant is moving on a plane horizontal surface. The number of degrees of freedom of the ant will be .........

Position of an ant ($S$ in metres) moving in $Y-Z$ plane is given by $S = 2t^2 \hat{j} + 5 \hat{k}$ (where $t$ is in seconds). The magnitude and direction of velocity of the ant at $t = 1 \ s$ will be:

The position vector of a moving body at any instant of time is given as $\overrightarrow{r} = (5t^2 \hat{i} - 5t \hat{j}) \text{ m}$. The magnitude and direction of velocity at $t = 2 \text{ s}$ is,

$A$ point moves in the $x-y$ plane according to $x = kt$ and $y = kt(1 - \alpha t)$,where $k$ and $\alpha$ are positive constants. The equation of the trajectory is:

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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 $t = 1$?