# Motion in a Plane - Class 11 - Physics

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## Extra Questions - Motion in a Plane | NCERT | Physics | Class 11

Airplanes $A$ and $B$ are in the same vertical plane and flying with constant velocity. At time $t = 0$ sec, an observer in $A$ finds $B$ at a distance of $500$ m and moving perpendicular to the line of motion of $A$. If at $t = t_{0}$ sec, $A$ just escapes being hit by $B$, then $t_{0}$ in seconds is:

To solve this problem, we consider the velocities of airplanes $A$ and $B$ and their directions of motion.

Given that:

Airplane $B$ is observed to be moving perpendicular to Airplane $A$'s line of motion.

Consider the plane of motion with x-axis along $A$'s trajectory and y-axis perpendicular to it. Assuming airplane $A$ is stationary for relative motion analysis, let's compute:

**Relative velocity**of $B$ with respect to $A$ along x-axis: $ (v_{BA}){ox} = (v_B){ox} - (v_A){ox} = 0 $

We know that $(v_A){ox} = 100\sqrt{3}$ m/s (assuming the escape speed given indirectly).

Assuming $(v_B)_{ox} = v \cos 30^\circ$, we equate: $ v \cos 30^\circ = 100 \sqrt{3} \Rightarrow v = 200 , \text{m/s} $ because $\cos 30^\circ = \frac{\sqrt{3}}{2}$.**Perpendicular distance**$d$ to be covered by $B$ to reach the path of $A$ is given as 500 m. Thus, the time needed for $B$ to travel this distance at the velocity perpendicular to $A$ is calculated using: $ t = \frac{d}{v_{oy}} $ where $v_{oy}$ represents velocity of $B$ perpendicular to $A$, which is $v \sin 30^\circ = 200 \times \frac{1}{2} = 100 , \text{m/s}$. Thus, $ t = \frac{500}{100} = 5 , \text{sec} $

**Conclusion:**The time $t_0$ at which airplane $A$ just avoids collision with airplane $B$ is **$t_0 = 5$ seconds**. This calculation uses the **components of velocity** and **geometry** of the motion to determine the narrow escape of $A$.

A player kicks a football at an angle of $45^{\circ}$ with a velocity of $30 \mathrm{~ms}^{-1}$. A second player 120 m away along the direction of the kick starts running to receive the ball at that instant. Find the speed with which the second player should run to reach the ball before it hits the ground ($\mathrm{g}=10 \mathrm{~ms}^{-2}$).

A) $5 \sqrt{2} \mathrm{~m/s}$ B) $4 \sqrt{2} \mathrm{~m/s}$ C) $5 \mathrm{~m/s}$ D) $10 \sqrt{2} \mathrm{~m/s}$

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