(D) The position vector of the ant is given by $\vec{S} = 2t^2 \hat{j} + 5 \hat{k}$.Velocity $\vec{v}$ is the time derivative of the position vector:

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(D) The position vector of the ant is given by $\vec{S} = 2t^2 \hat{j} + 5 \hat{k}$.Velocity $\vec{v}$ is the time derivative of the position vector: $\vec{v} = \frac{d\vec{S}}{dt} = \frac{d}{dt}(2t^2 \hat{j} + 5 \hat{k}) = 4t \hat{j}$.At $t = 1 \ s$,the velocity is $\vec{v} = 4(1) \hat{j} = 4 \hat{j} \ m/s$.The magnitude of velocity is $|\vec{v}| = 4 \ m/s$.The direction is along the positive $y$-axis (indicated by the unit vector $\hat{j}$).

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

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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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A
$16 \ m/s$ in $y$-direction
B
$4 \ m/s$ in $x$-direction
C
$9 \ m/s$ in $z$-direction
D
$4 \ m/s$ in $y$-direction

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3-2.Motion in Plane

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Motion In Two And Three Dimension

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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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$A$ 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.

$A$ particle moves in space along the path $z = ax^3 + by^2$ in such a way that $\frac{dx}{dt} = c = \frac{dy}{dt}$,where $a, b,$ and $c$ are constants. The acceleration of the particle is:

The $x$ and $y$ coordinates of a particle at any time $t$ are given by $x = 7t + 4t^2$ and $y = 5t$,where $x$ and $y$ are in $m$ and $t$ is in $s$. The acceleration of the particle at $t = 5 \ s$ is ......... $m/s^2$.

The coordinates of a moving particle at any time $t$ are given by $x = at^2$ and $y = bt^2$. The speed of the particle at any moment is

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