Question

Particle-1 is projected from ground (take it origin) at time $t=0$, with velocity $(30\hat{i}+30\hat{j})\,m{{s}^{-1}}$. Particle-2 is projected from $(130m,\,75m)$ at time $t=1s$ with velocity $(-20\hat{i}+20\hat{j})\,m{{s}^{-1}}$ . Assuming $\hat{j}$ to be vertically upwards and $\hat{i}$ to be in horizontal direction, match the following two columns at $t=2s$
List-1

A	Horizontal distance between two
B	Vertical distance between two
C	Relative horizontal component of velocity between two
D	Relative vertical component of velocity between two

 List-2

1	$60m{{s}^{-1}}$
2	$60m$
3	$50m$
4	$10m{{s}^{-1}}$

Answer 1

Hint: Two particles are moving in an x-y plane. Assume they are moving with constant velocities. To calculate the distance between the, calculate the difference between their positions. Relative velocity of one particle with respect to the other is the difference in their respective velocity components.
Formulas Used:
${{v}_{x}}=\dfrac{d}{t}$
${{v}_{y}}=\dfrac{d}{t}$
${{v}_{r}}={{v}_{1}}-{{v}_{2}}$

Complete step-by-step solution:
Given that, particle-1 is at $(0,0)$ at $t=0$, its component of velocity along the horizontal axis is $30\,m{{s}^{-1}}$. This means it is moving with $30\,m{{s}^{-1}}$ along the positive x-axis. Distance covered by it in $2s$ is-
${{v}_{x}}=\dfrac{d}{t}$
Here,$v$ is the velocity
$d$ is distance covered along x-axis (horizontal axis)
$t$ is time taken
Substituting values in the above equation, we get,
$\begin{align}
  & 30=\dfrac{d}{2} \\
 & \therefore d=60m \\
\end{align}$
 Particle-1 covers $60m$ in the positive x-direction. It is at position $x=60$.
 Given that, particle-2 is at $(130m,\,75m)$ at $t=1s$. Its component of velocity along the x-axis is $-20m{{s}^{-1}}$. This means that it is moving with $20m{{s}^{-1}}$ in the negative x-direction. Distance covered by it in $t=2-1=1s$ is given by-
${{v}_{x}}=\dfrac{d}{t}$
Substituting given values in the above equation, we get,
$\begin{align}
  & 20=\dfrac{d}{t} \\
 & \therefore d=20m \\
\end{align}$
 Particle-2 covers a distance of $20m$ in the negative x-direction. Its position at $t=2s$ is $(130-20)=110m$.
Horizontal distance between both = $110-60=50m$ - (1)
Similarly, for particle-1, vertical component of velocity is $30\,m{{s}^{-1}}$ distance covered along y-axis at $t=2s$ is given by-
${{v}_{y}}=\dfrac{d}{t}$ - (2)
${{v}_{y}}$is the vertical component of velocity
Substituting given values, we get,
$\begin{align}
  & 30=\dfrac{d}{2} \\
 & \therefore d=60m \\
\end{align}$
 Particle-1 covers $60m$ in the vertical direction. Its position is $y=60$.
For particle-2, the vertical component of its velocity is $20m{{s}^{-1}}$ . From eq (2), distance covered along y-axis in $t=2-1=1s$ is given by-
$\begin{align}
  & 20=\dfrac{d}{1} \\
 & \therefore d=20m \\
\end{align}$
Particle-2 covers$20m$in the vertical direction. Its position is
$\begin{align}
  & y=75+20 \\
 & y=95m \\
\end{align}$
Distance between both along y-axis= $95-60=35m$ - (3)
Particle-1 and particle-2 are moving in opposite directions; therefore their relative velocity will be-
${{v}_{r}}={{v}_{1}}-{{v}_{2}}$
 Here,
${{v}_{r}}$ is relative velocity along horizontal direction
${{v}_{1}}$ is velocity of particle-1
${{v}_{2}}$ is velocity of particle-2
Substituting values in the above equation, we get,
$\begin{align}
  & {{v}_{r}}=30-(-20) \\
 & {{v}_{r}}=50m{{s}^{-1}} \\
\end{align}$
Relative velocity between both along x-axis= $50m{{s}^{-1}}$ - (4)
Along y-axis,
$\begin{align}
  & {{v}_{r}}'={{v}_{1}}'-{{v}_{2}}' \\
 & {{v}_{r}}'=30-20=10m{{s}^{-1}} \\
\end{align}$
 Therefore, relative velocity along y-axis= $10m{{s}^{-1}}$ - (5)

The columns are matched as follows-

A	2
B	3
C	1
D	4

Note:
The negative sign indicates particles moving along the negative x-axis. To calculate relative velocity for two objects moving opposite each other, the velocities of both objects get added. The distance between two vectors is the difference between their positions in the plane.