# General Motion in a Plane

Motion in a plane refers to the movement of an object on a flat surface. The motion can be rectilinear, in which the object moves in a straight line, or curvilinear, in which the object moves in a curved path. It is described in a 2D coordinate system by using vectors (displacement, velocity, acceleration). The projectile motion and circular motion are examples of motion in a plane.

Let a particle moves in a plane. Consider a reference frame with origin O and orthogonal axes $x$ and $y$ in this plane. The location of O and the directions of $x$ and $y$ do not change with time. Let the particle be at a point $\mathrm{P}_1$ at time $t_1$. The position of the particle at time $t_1$ is specified by the position vector \begin{align} \vec{\mathrm{OP}}_1=\vec{r}_1=x_1\,\hat\imath+y_1\,\hat\jmath. \end{align}

Let the particle moves to the point $\mathrm{P}_2$ at time $t_2=t_1+\Delta t$. The position of the particle at time $t_2$ is \begin{align} \vec{\mathrm{OP}}_2=\vec{r}_2=x_2\,\hat\imath+y_2\,\hat\jmath. \end{align} The displacement of the particle in the time interval $t_1$ to $t_2$ is defined as \begin{align} \Delta\vec{r} &=\vec{r_2}-\vec{r_1} \\ &=(x_2-x_1)\hat\imath+(y_2-y_1)\hat\jmath.\nonumber \end{align} The average velocity of the particle in the time interval $t_1$ to $t_2$ is given by \begin{align} \vec{v}_\text{av}&=\frac{\Delta\vec{r}}{\Delta t} \\ &=\frac{x_2-x_1}{\Delta t}\hat\imath+\frac{y_2-y_1}{\Delta t}\hat\jmath \\ &=v_{\text{av,x}}\hat\imath+v_{\text{av,y}}\hat\jmath,\nonumber \end{align} where $v_{\text{av,x}}$ and $v_{\text{av,y}}$ are the average speeds along $x$ and $y$ directions.

The velocity (or instantaneous velocity) of the particle at a time $t$ is defined as \begin{align} \vec{v} &=\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \\ &=\frac{\mathrm{d}x}{\mathrm{d}t}\hat\imath+\frac{\mathrm{d}y}{\mathrm{d}t}\hat\jmath \\ &=v_x\hat\imath+v_y\hat\jmath.\nonumber \end{align}

Let the velocity of the particle changes from $\vec{v}_1$ to $\vec{v}_2$ in the time interval $t_1$ to $t_2$. The average acceleration of the particle in the time interval $t_1$ to $t_2$ is defined as \begin{align} \vec{a}_\text{av}&=\frac{\Delta\vec{v}}{\Delta t} \\ &=\frac{v_{2,x}-v_{1,x}}{\Delta t}\hat\imath+\frac{v_{2,y}-v_{1,y}}{\Delta t}\hat\jmath\nonumber\\ &=a_{\text{av,x}}\hat\imath+a_{\text{av,y}}\hat\jmath,\nonumber \end{align} where $a_{\text{av,x}}$ and $a_{\text{av,y}}$ are the average accelerations along $x$ and $y$ directions.

The acceleration of the particle at time $t$ is defined as \begin{align} \vec{a} &=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} \\ &=\frac{\mathrm{d}{v_x}}{\mathrm{d}t}\hat\imath+\frac{\mathrm{d}{v_y}}{\mathrm{d}t}\hat\jmath.\nonumber \end{align}

## Constant acceleration

Let a particle starts its motion from the origin at time $t=0$ with a velocity $\vec{u}$. If acceleration $\vec{a}$ of the particle is constant then its velocity $\vec{v}$ and position $\vec{r}$ at time $t$ are given by \begin{align} \vec{v}&=\vec{u}+\vec{a}\,t,\nonumber\\ \vec{r}&=\vec{u}\,t+\tfrac{1}{2}\vec{a}\,t^2.\nonumber \end{align} The average velocity of the particle is given by \begin{align} \vec{v}_\text{av}=(\vec{u}+\vec{v})/2. \end{align} The motion in a plane can be treated as two simultaneous rectilinear motions in orthogonal directions.

## Problems from IIT JEE

Problem (IIT JEE 2014): Airplanes A and B are flying with constant velocity in the same vertical plane at angles $30^\mathrm{o}$ and $60^\mathrm{o}$ with respect to the horizontal respectively as shown in figure.

The speed of A is $100\sqrt{3}\;\mathrm{m/s}$. At time $t={0}\;\mathrm{s}$, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at $t=t_0$, A just escapes being hit by B, $t_0$ in seconds is_________.

Solution: Let $\vec{V}_\text{A}$ and $\vec{V}_\text{B}$ be the velocity vectors of airplane A and B in a frame attached to the ground. The figure shows $\vec{V}_\text{A}$, $\vec{V}_\text{B}$, and $\vec{V}_\text{B/A}$, the velocity of B relative to A.

The expressions for these three vectors are, \begin{align} \vec{V}_\text{A} & =100\sqrt{3}\cos30\,\hat{\imath}+100\sqrt{3} \sin30\,\hat{\jmath} \\ &=150\,\hat{\imath}+50\sqrt{3}\,\hat{\jmath}, \\ \vec{V}_\text{B} &=V_\text{B}\cos60\,\hat\imath+V_\text{B}\sin60\,\hat\jmath \\ &=\frac{V_\text{B}}{2}\,\hat\imath+\frac{\sqrt{3}V_\text{B}}{2}\,\hat\jmath, \\ \vec{V}_\text{B/A} &=\vec{V}_\text{B}-\vec{V}_\text{A} \\ &=\left(\frac{V_\text{B}}{2}-150\right)\hat\imath+\sqrt{3}\left(\frac{V_\text{B}}{2}-50\right)\hat\jmath. \end{align} The observer in A sees B moving with a constant velocity perpendicular to the line of motion of A i.e., $\vec{V}_\text{B/A}\perp \vec{V}_\text{A}$. Thus, \begin{align} &\vec{V}_\text{B/A}\cdot\vec{V}_\text{A}=150(V_\text{B}-200)=0. \end{align} Solve above equation to get $V_\text{B}={200}\;\mathrm{m/s}$. Substitute $V_\text{B}$ to get $\vec{V}_\text{B/A}=-50\hat\imath+50\sqrt{3}\hat\jmath$ and $|\vec{V}_\text{B/A}|={100}\;\mathrm{m/s}$. The time taken to travel a relative distance of 500 m with a relative speed of 100 m/s is $t_0=500/100={5}\;\mathrm{s}$.