Motion In Two And Three Dimension Questions in English

(D) The displacement of the fly is equal to the length of the space diagonal of the rectangular hall.
The formula for the space diagonal $d$ of a rectangular parallelepiped with dimensions $l$,$b$,and $h$ is given by:
$d = \sqrt{l^2 + b^2 + h^2}$
Given dimensions are $l = 10\,m$,$b = 12\,m$,and $h = 14\,m$.
Substituting these values into the formula:
$d = \sqrt{10^2 + 12^2 + 14^2}$
$d = \sqrt{100 + 144 + 196}$
$d = \sqrt{440}$
$d \approx 20.97\,m$
Rounding to the nearest whole number,we get $d \approx 21\,m$.
Therefore,the correct option is $D$.

(A) Let the initial position be the origin $(0, 0, 0)$.
Moving $6\,m$ north corresponds to the $y$-axis,$8\,m$ east corresponds to the $x$-axis,and $10\,m$ vertically upwards corresponds to the $z$-axis.
The position vector of the body is $\vec{r} = 8\hat{i} + 6\hat{j} + 10\hat{k}$.
The magnitude of the resultant displacement is given by $r = |\vec{r}| = \sqrt{x^2 + y^2 + z^2}$.
Substituting the values: $r = \sqrt{8^2 + 6^2 + 10^2}$.
$r = \sqrt{64 + 36 + 100} = \sqrt{200}$.
$r = \sqrt{100 \times 2} = 10\sqrt{2}\,m$.

(A) Let the initial position be the origin $(0, 0, 0)$.
First,the aeroplane moves $400 \,m$ North (along $y$-axis) and $300 \,m$ South (along negative $y$-axis).
The net displacement in the horizontal plane is $400 \,m - 300 \,m = 100 \,m$ towards the North.
Then,it flies $1200 \,m$ upwards (along $z$-axis).
The final position vector is $\vec{r} = 100 \hat{j} + 1200 \hat{k}$.
The magnitude of the net displacement is $r = \sqrt{(100)^2 + (1200)^2}$.
$r = \sqrt{10000 + 1440000} = \sqrt{1450000}$.
$r = 100 \sqrt{145} \approx 100 \times 12.04 = 1204 \,m$.

(D) The velocity component along the $X$-axis is given by $v_x = \frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$.
The velocity component along the $Y$-axis is given by $v_y = \frac{dy}{dt} = \frac{d}{dt}(bt^2) = 2bt$.
The magnitude of the velocity (speed) of the particle is $v = \sqrt{v_x^2 + v_y^2}$.
Substituting the values,we get $v = \sqrt{(2at)^2 + (2bt)^2} = \sqrt{4a^2t^2 + 4b^2t^2} = \sqrt{4t^2(a^2 + b^2)}$.
Simplifying this,we get $v = 2t\sqrt{a^2 + b^2}$.

(C) The velocity components are given by the time derivatives of position:
$v_x = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 6t) = 6t - 6 \text{ m/s}$.
$v_y = \frac{dy}{dt} = \frac{d}{dt}(t^2 - 2t) = 2t - 2 \text{ m/s}$.
At $t = 1 \text{ s}$,$v_x = 6(1) - 6 = 0 \text{ m/s}$ and $v_y = 2(1) - 2 = 0 \text{ m/s}$.
Since both components are zero,the magnitude of velocity $v = \sqrt{v_x^2 + v_y^2} = 0 \text{ m/s}$ at $t = 1 \text{ s}$.
Therefore,the velocity of the particle is zero at $t = 1 \text{ s}$.

(B) The acceleration components are given by the second derivative of position with respect to time.
For the $x$-coordinate: $x = 7t + 4t^2$.
Velocity $v_x = \frac{dx}{dt} = 7 + 8t$.
Acceleration $a_x = \frac{dv_x}{dt} = 8 \ m/s^2$.
For the $y$-coordinate: $y = 5t$.
Velocity $v_y = \frac{dy}{dt} = 5$.
Acceleration $a_y = \frac{dv_y}{dt} = 0 \ m/s^2$.
The net acceleration is $a = \sqrt{a_x^2 + a_y^2} = \sqrt{8^2 + 0^2} = 8 \ m/s^2$.
Since the acceleration is constant,it remains $8 \ m/s^2$ at $t = 5 \ s$.

(C) Given the position coordinates of the particle as $x = \alpha t^3$ and $y = \beta t^3$.
The velocity components are obtained by differentiating the position with respect to time $t$:
$v_x = \frac{dx}{dt} = \frac{d}{dt}(\alpha t^3) = 3\alpha t^2$
$v_y = \frac{dy}{dt} = \frac{d}{dt}(\beta t^3) = 3\beta t^2$
The speed $v$ of the particle is the magnitude of the velocity vector:
$v = \sqrt{v_x^2 + v_y^2}$
$v = \sqrt{(3\alpha t^2)^2 + (3\beta t^2)^2}$
$v = \sqrt{9\alpha^2 t^4 + 9\beta^2 t^4}$
$v = \sqrt{9t^4(\alpha^2 + \beta^2)}$
$v = 3t^2\sqrt{\alpha^2 + \beta^2}$

(D) The average velocity vector between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}}{\Delta t}$.
The slope of the secant line connecting these two points is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
For the points $A(0, 2)$ and $B(5, 3)$,the slope is $m = \frac{3 - 2}{5 - 0} = \frac{1}{5} = 0.2$.
The instantaneous velocity at any point is given by the slope of the tangent to the trajectory at that point,$v_y / v_x = dy/dx$.
We are looking for a point where the tangent slope is equal to the secant slope,i.e.,$dy/dx = 0.2$.
Looking at the graph,at the point $(3, 1)$,the curve is transitioning from a downward slope to an upward slope,and the tangent is horizontal $(slope = 0)$.
At the point $(4, 2)$,the tangent slope is positive and visually matches the slope of the line connecting $(0, 2)$ and $(5, 3)$.
Thus,the correct coordinates are $(4, 2)$.

(C) Given that the body starts from rest,the initial velocity components are $u_x = 0$ and $u_y = 0$.
The acceleration components are $a_x = 6 \; m/s^2$ and $a_y = 8 \; m/s^2$.
The time elapsed is $t = 4 \; s$.
Using the kinematic equation $s = ut + \frac{1}{2}at^2$ for both axes:
For the $x$-axis: $x = 0(4) + \frac{1}{2} \times 6 \times (4)^2 = 3 \times 16 = 48 \; m$.
For the $y$-axis: $y = 0(4) + \frac{1}{2} \times 8 \times (4)^2 = 4 \times 16 = 64 \; m$.
The distance from the origin is given by $d = \sqrt{x^2 + y^2}$.
$d = \sqrt{(48)^2 + (64)^2} = \sqrt{2304 + 4096} = \sqrt{6400} = 80 \; m$.

(A) Given the position coordinates of the particle as functions of time $t$:
$x = 5t - 2t^2$
$y = 10t$
To find the velocity components,we differentiate the position coordinates with respect to time $t$:
$v_x = \frac{dx}{dt} = \frac{d}{dt}(5t - 2t^2) = 5 - 4t$
$v_y = \frac{dy}{dt} = \frac{d}{dt}(10t) = 10$
To find the acceleration components,we differentiate the velocity components with respect to time $t$:
$a_x = \frac{dv_x}{dt} = \frac{d}{dt}(5 - 4t) = -4 \, m/s^2$
$a_y = \frac{dv_y}{dt} = \frac{d}{dt}(10) = 0 \, m/s^2$
The acceleration vector is $\vec{a} = a_x \hat{i} + a_y \hat{j} = -4 \hat{i} + 0 \hat{j} = -4 \hat{i} \, m/s^2$.
Since the acceleration components are constant,the acceleration of the particle at any time $t$,including $t = 2 \, s$,is $-4 \, m/s^2$ in the $x$-direction.

(C) The average velocity vector $\vec{v}_{avg}$ from the starting point $P$ to any point $Q$ on the path is given by $\vec{v}_{avg} = \frac{\vec{r}_Q - \vec{r}_P}{t_Q - t_P}$,which is the direction of the displacement vector $\Delta \vec{r} = \vec{r}_Q - \vec{r}_P$.
The instantaneous velocity vector $\vec{v}_{inst}$ at any point is tangent to the path at that point.
For the average velocity vector to be in the same direction as the instantaneous velocity vector,the displacement vector $\vec{r}_Q - \vec{r}_P$ must be tangent to the path at point $Q$.
Geometrically,this means the line segment connecting the starting point $P$ to point $Q$ must be tangent to the curve at point $Q$.
Looking at the path,if we draw a line from $P$ to $C$,this line is tangent to the path at point $C$. Therefore,at point $C$,the average velocity vector is in the same direction as the instantaneous velocity vector.

(C) Given velocity $\vec{v} = v_x \hat{i} + v_y \hat{j} = a \hat{i} + bx \hat{j}$.
Thus,$v_x = \frac{dx}{dt} = a$ and $v_y = \frac{dy}{dt} = bx$.
From $v_x = a$,we integrate with respect to time: $\int_0^x dx = \int_0^t a dt$,which gives $x = at$,or $t = \frac{x}{a}$.
Now,substitute $t$ into the expression for $v_y$: $\frac{dy}{dt} = bx$.
Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{bx}{a}$.
Integrating both sides with respect to $x$: $\int_0^y dy = \int_0^x \frac{b}{a} x dx$.
This yields $y = \frac{b}{a} \cdot \frac{x^2}{2} = \frac{bx^2}{2a}$.

(A) Given the equations of motion:
$x = kt$
$y = kt(1 - \alpha t)$
From the first equation,we can express time $t$ as:
$t = \frac{x}{k}$
Substitute this value of $t$ into the equation for $y$:
$y = k \left( \frac{x}{k} \right) \left( 1 - \alpha \left( \frac{x}{k} \right) \right)$
$y = x \left( 1 - \frac{\alpha x}{k} \right)$
$y = x - \frac{\alpha x^2}{k}$
Thus,the equation of the trajectory is $y = x - \frac{\alpha x^2}{k}$.

(A) Given that $\frac{dx}{dt} = c$ and $\frac{dy}{dt} = c$. Since $c$ is a constant,the second derivatives are $\frac{d^2x}{dt^2} = 0$ and $\frac{d^2y}{dt^2} = 0$.
The path is given by $z = ax^3 + by^2$.
To find the velocity component in the $z$-direction,differentiate with respect to time $t$:
$\frac{dz}{dt} = 3ax^2 \frac{dx}{dt} + 2by \frac{dy}{dt} = 3ax^2(c) + 2by(c) = 3acx^2 + 2bcy$.
To find the acceleration component in the $z$-direction,differentiate $\frac{dz}{dt}$ with respect to time $t$:
$\frac{d^2z}{dt^2} = \frac{d}{dt}(3acx^2 + 2bcy) = 3ac(2x \frac{dx}{dt}) + 2bc(\frac{dy}{dt})$.
Substituting $\frac{dx}{dt} = c$ and $\frac{dy}{dt} = c$:
$\frac{d^2z}{dt^2} = 6acx(c) + 2bc(c) = 6ac^2x + 2bc^2$.
The acceleration vector is $\vec{a} = \frac{d^2x}{dt^2} \hat{i} + \frac{d^2y}{dt^2} \hat{j} + \frac{d^2z}{dt^2} \hat{k}$.
Substituting the values,we get $\vec{a} = 0 \hat{i} + 0 \hat{j} + (6ac^2x + 2bc^2) \hat{k} = (6ac^2x + 2bc^2) \hat{k}$.

(A) The velocity components are obtained by differentiating the position coordinates with respect to time $t$:
$v_x = \frac{dx}{dt} = -a \omega \sin \omega t$
$v_y = \frac{dy}{dt} = a \omega \cos \omega t$
$v_z = \frac{dz}{dt} = a \omega$
The speed $v$ is the magnitude of the velocity vector:
$v = \sqrt{v_x^2 + v_y^2 + v_z^2}$
$v = \sqrt{(-a \omega \sin \omega t)^2 + (a \omega \cos \omega t)^2 + (a \omega)^2}$
$v = \sqrt{a^2 \omega^2 \sin^2 \omega t + a^2 \omega^2 \cos^2 \omega t + a^2 \omega^2}$
$v = \sqrt{a^2 \omega^2 (\sin^2 \omega t + \cos^2 \omega t) + a^2 \omega^2}$
Since $\sin^2 \omega t + \cos^2 \omega t = 1$,we get:
$v = \sqrt{a^2 \omega^2 (1) + a^2 \omega^2} = \sqrt{2 a^2 \omega^2} = \sqrt{2} a \omega$

(D) The position vector is given by $\vec{r}(t) = (15t^2) \hat{i} + (4 - 20t^2) \hat{j}$.
To find the velocity vector $\vec{v}$,we differentiate $\vec{r}$ with respect to time $t$:
$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(15t^2) \hat{i} + \frac{d}{dt}(4 - 20t^2) \hat{j} = (30t) \hat{i} - (40t) \hat{j}$.
To find the acceleration vector $\vec{a}$,we differentiate $\vec{v}$ with respect to time $t$:
$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(30t) \hat{i} - \frac{d}{dt}(40t) \hat{j} = 30 \hat{i} - 40 \hat{j}$.
The acceleration is constant and independent of time.
The magnitude of the acceleration is $|\vec{a}| = \sqrt{(30)^2 + (-40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \ m/s^2$.

(N/A) $v(t) = \frac{dr}{dt} = \frac{d}{dt}(3.0 t \hat{i} + 2.0 t^{2} \hat{j} + 5.0 \hat{k}) = 3.0 \hat{i} + 4.0 t \hat{j} \ m/s$.
$a(t) = \frac{dv}{dt} = \frac{d}{dt}(3.0 \hat{i} + 4.0 t \hat{j}) = 4.0 \hat{j} \ m/s^{2}$.
At $t = 1.0 \ s$,the velocity vector is $v = 3.0 \hat{i} + 4.0 \hat{j} \ m/s$.
The magnitude is $|v| = \sqrt{3.0^{2} + 4.0^{2}} = \sqrt{9 + 16} = 5.0 \ m/s$.
The direction $\theta$ with the $x$-axis is given by $\theta = \tan^{-1}(\frac{v_{y}}{v_{x}}) = \tan^{-1}(\frac{4.0}{3.0}) \approx 53^{\circ}$.

(A) The position vector of the particle at time $t$ is given by $r(t) = v_0 t + \frac{1}{2} a t^2$.
Given $v_0 = 5.0 \hat{i} \; m/s$ and $a = (3.0 \hat{i} + 2.0 \hat{j}) \; m/s^2$,we have:
$r(t) = (5.0 \hat{i})t + \frac{1}{2}(3.0 \hat{i} + 2.0 \hat{j})t^2 = (5.0t + 1.5t^2) \hat{i} + (1.0t^2) \hat{j}$.
Thus,$x(t) = 5.0t + 1.5t^2$ and $y(t) = 1.0t^2$.
For $x = 84 \; m$,we solve $1.5t^2 + 5.0t - 84 = 0$. Using the quadratic formula,$t = \frac{-5.0 \pm \sqrt{25 + 4(1.5)(84)}}{2(1.5)} = \frac{-5.0 \pm \sqrt{529}}{3} = \frac{-5.0 \pm 23}{3}$. Since $t > 0$,$t = 6 \; s$.
$(a)$ At $t = 6 \; s$,$y = 1.0(6)^2 = 36.0 \; m$.
$(b)$ The velocity vector is $v(t) = \frac{dr}{dt} = (5.0 + 3.0t) \hat{i} + (2.0t) \hat{j}$.
At $t = 6 \; s$,$v = (5.0 + 3.0(6)) \hat{i} + (2.0(6)) \hat{j} = 23.0 \hat{i} + 12.0 \hat{j} \; m/s$.
The speed is $|v| = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673} \approx 25.94 \; m/s \approx 26 \; m/s$.

(N/A) The position of the particle is given by:
$\vec{r} = 3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}$
Velocity $\vec{v}$ is the time derivative of position:
$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}) = 3.0 \hat{i} - 4.0 t \hat{j} \; m/s$
Acceleration $\vec{a}$ is the time derivative of velocity:
$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(3.0 \hat{i} - 4.0 t \hat{j}) = -4.0 \hat{j} \; m/s^{2}$
$(b)$ At $t = 2.0 \; s$,the velocity vector is:
$\vec{v} = 3.0 \hat{i} - 4.0(2.0) \hat{j} = 3.0 \hat{i} - 8.0 \hat{j} \; m/s$
The magnitude is $|\vec{v}| = \sqrt{(3.0)^{2} + (-8.0)^{2}} = \sqrt{9 + 64} = \sqrt{73} \approx 8.54 \; m/s$
The direction $\theta$ with respect to the $x$-axis is:
$\theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) = \tan^{-1}\left(\frac{-8.0}{3.0}\right) \approx -69.45^{\circ}$
The negative sign indicates the direction is $69.45^{\circ}$ below the positive $x$-axis.

(A) Given: Initial velocity $\vec{u} = 10.0 \hat{j} \; m/s$,Acceleration $\vec{a} = (8.0 \hat{i} + 2.0 \hat{j}) \; m/s^2$.
Using the kinematic equation for position: $\vec{r}(t) = \vec{u}t + \frac{1}{2} \vec{a}t^2$.
Substituting the values: $\vec{r}(t) = (10.0 \hat{j})t + \frac{1}{2} (8.0 \hat{i} + 2.0 \hat{j})t^2 = 4.0 t^2 \hat{i} + (10t + t^2) \hat{j}$.
Equating components: $x = 4.0 t^2$ and $y = 10t + t^2$.
For $x = 16 \; m$: $16 = 4.0 t^2 \implies t^2 = 4 \implies t = 2 \; s$.
At $t = 2 \; s$,$y = 10(2) + (2)^2 = 20 + 4 = 24 \; m$.
$(b)$ Velocity vector $\vec{v}(t) = \vec{u} + \vec{a}t = 10.0 \hat{j} + (8.0 \hat{i} + 2.0 \hat{j})t = 8.0 t \hat{i} + (10 + 2.0 t) \hat{j}$.
At $t = 2 \; s$: $\vec{v} = 8.0(2) \hat{i} + (10 + 2.0(2)) \hat{j} = 16 \hat{i} + 14 \hat{j}$.
Speed $|\vec{v}| = \sqrt{16^2 + 14^2} = \sqrt{256 + 196} = \sqrt{452} \approx 21.26 \; m/s$.

(B) In one-dimensional motion,physical quantities like displacement,velocity,and acceleration can be described using $(+)$ and $(-)$ signs to indicate direction,as there are only two possible directions along a straight line.
However,when an object moves in two dimensions (a plane) or three dimensions (space),the direction is no longer restricted to just two possibilities. In such cases,we must use vectors to fully describe these physical quantities,as vectors account for both magnitude and direction in any arbitrary orientation.

(N/A) One-dimensional motion: The motion of a train moving along a straight track is an example of one-dimensional motion,as it requires only one coordinate to specify its position at any time.
Two-dimensional motion: The motion of a striker on a carrom board is an example of two-dimensional motion,as the striker moves in a plane and requires two coordinates $(x, y)$ to specify its position.
Three-dimensional motion: The motion of a fish swimming in water or a bird flying in the sky is an example of three-dimensional motion,as it requires three coordinates $(x, y, z)$ to specify its position at any given time.

(N/A) Position vector: The position vector $\vec{r}$ of a particle $P$ located in a plane with reference to the origin $O$ is given by:
$\vec{r} = x \hat{i} + y \hat{j}$
Where $x$ and $y$ are the components of $\vec{r}$ along the $x$ and $y$-axes,respectively,representing the coordinates of the object.
Displacement vector:
Suppose a particle moves along a curve and is at position $P$ at time $t$ and at position $P^{\prime}$ at time $t^{\prime}$.
At $P$,the position vector is $\vec{r} = x \hat{i} + y \hat{j}$.
At $P^{\prime}$,the position vector is $\vec{r}^{\prime} = x^{\prime} \hat{i} + y^{\prime} \hat{j}$.
The displacement vector $\Delta \vec{r}$ is the change in the position vector from $P$ to $P^{\prime}$:
$\Delta \vec{r} = \vec{r}^{\prime} - \vec{r}$
$\Delta \vec{r} = (x^{\prime} - x) \hat{i} + (y^{\prime} - y) \hat{j}$
$\Delta \vec{r} = \Delta x \hat{i} + \Delta y \hat{j}$
Where $\Delta x = x^{\prime} - x$ and $\Delta y = y^{\prime} - y$ are the changes in the $x$ and $y$ coordinates,respectively.

(N/A) The average velocity $(\vec{v})$ of an object is the ratio of the displacement and the corresponding time interval.
Suppose an object covers a displacement $\Delta \vec{r}$ in a time interval $\Delta t$.
Average velocity is given by:
$\langle\vec{v}\rangle = \frac{\Delta \vec{r}}{\Delta t} = \frac{\Delta x \hat{i} + \Delta y \hat{j}}{\Delta t} = \hat{i} \left( \frac{\Delta x}{\Delta t} \right) + \hat{j} \left( \frac{\Delta y}{\Delta t} \right)$
Or,$\langle\vec{v}\rangle = \langle v_{x} \rangle \hat{i} + \langle v_{y} \rangle \hat{j}$
The direction of the average velocity is the same as that of the displacement vector $\Delta \vec{r}$.
Instantaneous velocity is defined as the limiting value of the average velocity as the time interval approaches zero:
$\vec{v} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \vec{r}}{\Delta t} = \frac{d\vec{r}}{dt}$
The direction of velocity at any point on the path is tangential to the path at that point and is in the direction of motion.
In terms of components,the instantaneous velocity is:
$\vec{v} = \hat{i} \left( \frac{dx}{dt} \right) + \hat{j} \left( \frac{dy}{dt} \right) = v_{x} \hat{i} + v_{y} \hat{j}$
where $v_{x} = \frac{dx}{dt}$ and $v_{y} = \frac{dy}{dt}$ are the components of velocity along the $x$ and $y$ axes respectively.
The magnitude of the velocity vector is given by:
$v = \sqrt{v_{x}^{2} + v_{y}^{2}}$
The units of velocity in the $MKS$ system are $m/s$ and in the $CGS$ system are $cm/s$.

Average acceleration is defined as the time rate of change of velocity over a given time interval.
$\text{Average acceleration} = \frac{\text{Change in velocity}}{\text{Time interval}}$
The average acceleration $\vec{a}$ of an object moving in the $xy$-plane for a time interval $\Delta t$ is the change in velocity divided by the time interval:
$\vec{a} = \frac{\overrightarrow{\Delta v}}{\Delta t} = \frac{\Delta(v_x \hat{i} + v_y \hat{j})}{\Delta t} = \frac{\Delta v_x}{\Delta t} \hat{i} + \frac{\Delta v_y}{\Delta t} \hat{j} = a_x \hat{i} + a_y \hat{j}$
Instantaneous acceleration is the limiting value of the average acceleration as the time interval approaches zero:
$\vec{a} = \lim_{\Delta t \rightarrow 0} \frac{\overrightarrow{\Delta v}}{\Delta t} = \frac{d\vec{v}}{dt}$
Since $\vec{v} = v_x \hat{i} + v_y \hat{j}$,we have:
$\vec{a} = \frac{d}{dt}(v_x \hat{i} + v_y \hat{j}) = \frac{dv_x}{dt} \hat{i} + \frac{dv_y}{dt} \hat{j} = a_x \hat{i} + a_y \hat{j}$
where $a_x = \frac{dv_x}{dt}$ and $a_y = \frac{dv_y}{dt}$.
Furthermore,since $\vec{v} = \frac{d\vec{r}}{dt}$,acceleration can be expressed as the second derivative of position with respect to time:
$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}\left(\frac{d\vec{r}}{dt}\right) = \frac{d^2\vec{r}}{dt^2} = \ddot{\vec{r}}$

(B) The position vector of the particle is given by $\vec{r}(t) = (3t)\hat{i} + (4t^2)\hat{j}$.
To find the velocity vector $\vec{v}(t)$,we differentiate the position vector with respect to time $t$:
$\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt} [(3t)\hat{i} + (4t^2)\hat{j}]$
$\vec{v}(t) = 3\hat{i} + (8t)\hat{j}$.
Now,substitute $t = 2 \ s$ into the velocity equation:
$\vec{v}(2) = 3\hat{i} + (8 \times 2)\hat{j}$
$\vec{v}(2) = 3\hat{i} + 16\hat{j} \ m/s$.

(D) In motion in two or three dimensions,the velocity vector $\vec{v}$ and acceleration vector $\vec{a}$ can have any angle $\theta$ between them such that $0^{\circ} \le \theta \le 180^{\circ}$.
If $\theta = 0^{\circ}$,the speed increases.
If $\theta = 180^{\circ}$,the speed decreases.
If $\theta = 90^{\circ}$,the speed remains constant (uniform circular motion).
Therefore,the angle can be any value in the range $[0^{\circ}, 180^{\circ}]$.

Consider a particle moving in a plane with constant acceleration $\vec{a}$. At time $t=0$,the velocity is initial velocity $\vec{v_0}$ and the position is $\vec{r_0}$. At time $t=t$,the velocity is $\vec{v}$ and the position is $\vec{r}$.
$1$. Derivation of $\vec{v} = \vec{v_0} + \vec{a}t$:
Since the acceleration is constant,the instantaneous acceleration is given by:
$\vec{a} = \frac{\vec{v} - \vec{v_0}}{t - 0}$
$\vec{a} = \frac{\vec{v} - \vec{v_0}}{t}$
$\vec{v} = \vec{v_0} + \vec{a}t$
In component form:
$v_x = v_{0x} + a_x t$
$v_y = v_{0y} + a_y t$
$2$. Derivation of $\vec{r} = \vec{r_0} + \vec{v_0}t + \frac{1}{2}\vec{a}t^2$:
For constant acceleration,the average velocity is $\vec{v}_{avg} = \frac{\vec{v} + \vec{v_0}}{2}$.
The displacement is $\vec{r} - \vec{r_0} = \vec{v}_{avg} \cdot t = \left( \frac{\vec{v} + \vec{v_0}}{2} \right) t$.
Substituting $\vec{v} = \vec{v_0} + \vec{a}t$:
$\vec{r} - \vec{r_0} = \left( \frac{\vec{v_0} + \vec{a}t + \vec{v_0}}{2} \right) t$
$\vec{r} - \vec{r_0} = \left( \frac{2\vec{v_0} + \vec{a}t}{2} \right) t$
$\vec{r} = \vec{r_0} + \vec{v_0}t + \frac{1}{2}\vec{a}t^2$.

(B) Motion in a plane is defined as motion in two dimensions.
According to the principle of independence of motion,any motion in a plane can be treated as the combination of two independent motions along two mutually perpendicular directions (typically the $x$ and $y$ axes).
Therefore,the correct option is $B$.

(N/A) For motion in a plane with uniform acceleration $\vec{a}$,the motion can be resolved into two independent components along the $x$ and $y$ axes.
$1$. For the $x$-axis:
$v_x = u_x + a_x t$
$x = u_x t + \frac{1}{2} a_x t^2$
$v_x^2 = u_x^2 + 2 a_x x$
$2$. For the $y$-axis:
$v_y = u_y + a_y t$
$y = u_y t + \frac{1}{2} a_y t^2$
$v_y^2 = u_y^2 + 2 a_y y$
Here,$\vec{u} = (u_x, u_y)$ is the initial velocity,$\vec{v} = (v_x, v_y)$ is the final velocity at time $t$,and $\vec{a} = (a_x, a_y)$ is the constant acceleration.

(C) Given the coordinates of the particle as functions of time $t$:
$x = 4t^2$
$y = 2t$
To find the locus (path) of the particle,we eliminate the parameter $t$ from the equations.
From the second equation,we have $t = \frac{y}{2}$.
Substituting this value of $t$ into the first equation:
$x = 4 \left( \frac{y}{2} \right)^2$
$x = 4 \left( \frac{y^2}{4} \right)$
$x = y^2$
The equation $x = y^2$ represents a parabola opening along the positive $x$-axis.
Therefore,the locus of the particle is a parabola.

(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:
$d = \sqrt{(6 - 3)^2 + (5 - 1)^2}$
$d = \sqrt{3^2 + 4^2}$
$d = \sqrt{9 + 16} = \sqrt{25} = 5$.

(B) The position vector of the particle at any time $t$ is given by $\vec{r}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k}$.
Given $x = 3t$,$y = 5t^3$,and $z = 7$ (constant).
So,$\vec{r}(t) = 3t \hat{i} + 5t^3 \hat{j} + 7 \hat{k}$.
The velocity vector $\vec{v}$ is the first derivative of position with respect to time: $\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3t) \hat{i} + \frac{d}{dt}(5t^3) \hat{j} + \frac{d}{dt}(7) \hat{k} = 3 \hat{i} + 15t^2 \hat{j}$.
The acceleration vector $\vec{a}$ is the derivative of velocity with respect to time: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(3) \hat{i} + \frac{d}{dt}(15t^2) \hat{j} = 0 \hat{i} + 30t \hat{j}$.
At $t = 1 \, \text{s}$,the acceleration is $\vec{a} = 30(1) \hat{j} = 30 \hat{j} \, \text{cm/s}^2$.

(B) The position vector is given by $\vec{r} = 3t^2 \hat{i} + 6t \hat{j} + \hat{k}$.
To find the velocity vector $\vec{v}$,we differentiate the position vector with respect to time $t$:
$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3t^2 \hat{i} + 6t \hat{j} + \hat{k})$
$\vec{v} = (6t) \hat{i} + (6) \hat{j} + (0) \hat{k}$
The velocity component along the $y$-axis is the coefficient of the $\hat{j}$ unit vector,which is $v_y = 6$.
Thus,the magnitude of the velocity along the $y$-axis is $6$.

(A) Given velocity vector: $\vec{v} = v_x \hat{i} + v_y \hat{j} = 3 \hat{i} + 6x \hat{j}$.
Comparing components,we get $v_x = \frac{dx}{dt} = 3$ and $v_y = \frac{dy}{dt} = 6x$.
The slope of the path is given by $\frac{dy}{dx} = \frac{v_y}{v_x}$.
Substituting the values,$\frac{dy}{dx} = \frac{6x}{3} = 2x$.
Integrating both sides with respect to $x$: $\int dy = \int 2x dx$.
Since the particle starts from the origin $(0,0)$,the constant of integration is $0$.
Thus,$y = x^2$.

(B) The number of degrees of freedom of a particle is defined as the number of independent coordinates required to specify its position in space.
Since the ant is moving on a plane horizontal surface,its position can be completely described by two coordinates,typically $(x, y)$ in a Cartesian coordinate system.
Therefore,the ant has $2$ degrees of freedom.

(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 $\hat{j}$).

(C) The position vector is given by $\overrightarrow{r} = 5t^2 \hat{i} - 5t \hat{j}$.
The velocity vector $\overrightarrow{v}$ is the derivative of position with respect to time: $\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt} = 10t \hat{i} - 5 \hat{j}$.
At $t = 2 \text{ s}$,the velocity vector is $\overrightarrow{v} = 10(2) \hat{i} - 5 \hat{j} = 20 \hat{i} - 5 \hat{j} \text{ m/s}$.
The magnitude of velocity is $|\overrightarrow{v}| = \sqrt{(20)^2 + (-5)^2} = \sqrt{400 + 25} = \sqrt{425} = \sqrt{25 \times 17} = 5\sqrt{17} \text{ m/s}$.
To find the direction,let $\theta$ be the angle with the negative $\text{Y}$-axis. From the vector components,$\tan \theta = \frac{|v_x|}{|v_y|} = \frac{20}{5} = 4$. Thus,$\theta = \tan^{-1} 4$ with the negative $\text{Y}$-axis.

(D) Given: Initial velocity $\vec{u} = 10 \hat{j} \text{ ms}^{-1}$, Acceleration $\vec{a} = 8 \hat{i} + 2 \hat{j} \text{ ms}^{-2}$.
Using the kinematic equation $\vec{s} = \vec{u}t + \frac{1}{2} \vec{a}t^2$:
$\vec{s} = (10 \hat{j})t + \frac{1}{2} (8 \hat{i} + 2 \hat{j})t^2$
$\vec{s} = (4t^2) \hat{i} + (10t + t^2) \hat{j}$.
Comparing the components with $\vec{s} = x \hat{i} + y \hat{j}$, we get $x = 4t^2$ and $y = 10t + t^2$.
Given $x = 16 \text{ m}$, so $4t^2 = 16 \Rightarrow t^2 = 4 \Rightarrow t = 2 \text{ s}$.
Substituting $t = 2 \text{ s}$ into the expression for $y$:
$y = 10(2) + (2)^2 = 20 + 4 = 24 \text{ m}$.

(B) The position vector of the particle is given by $\vec{r} = x\hat{i} + y\hat{j} = (4t + t^2)\hat{i} + (2t + \frac{t^2}{2})\hat{j}$.
Velocity is the time derivative of the position vector: $\vec{v} = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}$.
Differentiating $x$ with respect to $t$: $v_x = \frac{d}{dt}(4t + t^2) = 4 + 2t$.
Differentiating $y$ with respect to $t$: $v_y = \frac{d}{dt}(2t + \frac{t^2}{2}) = 2 + t$.
Therefore,the velocity vector is $\vec{v} = (4 + 2t)\hat{i} + (2 + t)\hat{j} \text{ m/s}$.

(A) Given the position vector: $\vec{r} = 3t^2 \hat{i} + 2t \hat{j} + \hat{k}$.
Velocity is the time derivative of position: $\vec{v} = \frac{d\vec{r}}{dt} = 6t \hat{i} + 2 \hat{j}$.
At $t = 2 \text{ s}$,the velocity vector is $\vec{v} = 6(2) \hat{i} + 2 \hat{j} = 12 \hat{i} + 2 \hat{j}$.
The magnitude of velocity is $|\vec{v}| = \sqrt{12^2 + 2^2} = \sqrt{144 + 4} = \sqrt{148} \text{ m/s}$.
Acceleration is the time derivative of velocity: $\vec{a} = \frac{d\vec{v}}{dt} = 6 \hat{i}$.
The magnitude of acceleration is $|\vec{a}| = \sqrt{6^2} = 6 \text{ m/s}^2$.
Thus,the magnitude of acceleration is $6$ and the magnitude of velocity is $\sqrt{148}$.

(C) The velocity vector is given by $\vec{V} = V_x \hat{i} + V_y \hat{j}$,where $V_x = 6 + 2t$ and $V_y = 4 + 2\sqrt{3}t$.
Acceleration $\vec{a}$ is the time derivative of velocity: $\vec{a} = \frac{d\vec{V}}{dt} = \frac{dV_x}{dt} \hat{i} + \frac{dV_y}{dt} \hat{j}$.
Calculating the components:
$a_x = \frac{d}{dt}(6 + 2t) = 2 \text{ m/s}^2$.
$a_y = \frac{d}{dt}(4 + 2\sqrt{3}t) = 2\sqrt{3} \text{ m/s}^2$.
Therefore,the acceleration vector is $\vec{a} = 2 \hat{i} + 2\sqrt{3} \hat{j} \text{ m/s}^2$.

(A) Given the parametric equations of motion:
$x = a \cos t$
$y = a \sin t$
$z = t$
First,consider the projection of the motion on the $xy$-plane:
$x^2 + y^2 = (a \cos t)^2 + (a \sin t)^2 = a^2(\cos^2 t + \sin^2 t) = a^2$
This represents a circle of radius $a$ in the $xy$-plane.
Simultaneously,the particle moves along the $z$-axis with a constant velocity since $z = t$,which implies $\frac{dz}{dt} = 1$.
Since the particle moves in a circular path in the $xy$-plane while simultaneously advancing linearly along the $z$-axis,the resulting trajectory is a helix.