Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30° and 60°

Q: Why do we subtract $30^\circ$ from $60^\circ$ to find the angle between the velocities?

Because both angles are measured from the same horizontal reference line. The angle between their paths is simply the difference between their individual angles with the horizontal.

Q: Why must the parallel component of $B$'s velocity equal $A$'s velocity?

If the observer in $A$ sees $B$ moving *only* perpendicularly, it means $B$ is not moving forwards or backwards relative to $A$. For this to happen, their speeds in the forward (parallel) direction must be exactly identical.

Q: What happens to the perpendicular component of $A$'s velocity?

Airplane $A$ is moving entirely along its own path, so it has no velocity component perpendicular to itself. Therefore, the relative perpendicular velocity is entirely due to airplane $B$.

Q: Why don't we need to know the exact starting coordinates of $A$ and $B$?

Because the problem gives us the initial *relative* distance ($500 \text{ m}$) and the direction of relative motion. In the relative frame, $B$ is simply moving along a straight line towards $A$'s path from $500 \text{ m}$ away.

Q: Could we solve this without relative velocity?

Yes, by writing the $x$ and $y$ position equations for both airplanes as functions of time and finding when their paths intersect. However, the relative velocity method turns a complex 2D intersection problem into a simple 1D constant-speed calculation!

Question

Airplanes $A$ and $B$ are flying with constant velocity in the same vertical plane at angles $30^\circ$ and $60^\circ$ with respect to the horizontal respectively as shown in figure. The speed of $A$ is $100\sqrt{3} 	ext{ ms}^{-1}$. At time $t = 0 	ext{ s}$, an observer in $A$ finds $B$ at a distance of $500 	ext{ m}$. 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

Sigma Percentile · Accepted Answer

### The Setup: Two Airplanes in the Sky

Imagine two airplanes, $A$ and $B$, soaring through the sky. Airplane $A$ is flying at an angle of $30^\circ$ to the horizontal with a speed of $100\sqrt{3} 	ext{ m/s}$. Meanwhile, airplane $B$ is flying at a steeper angle of $60^\circ$ to the horizontal.

At a specific moment, let's call it $t = 0$, an observer sitting in airplane $A$ spots airplane $B$ exactly $500 	ext{ m}$ away. But here is the fascinating part: to the observer in $A$, airplane $B$ appears to be moving in a straight line that is **perfectly perpendicular** to $A$'s own line of motion.

Our mission is to find out exactly when airplane $B$ will cross $A$'s path, a time denoted as $t_0$.

### The Power of Relative Velocity

When a problem involves two moving bodies and an observer on one of them, it is a massive hint to use the concept of **relative velocity**. Instead of tracking both airplanes from the ground, we shift our frame of reference to airplane $A$.

In this new frame, airplane $A$ is considered stationary, and we only care about the velocity of $B$ with respect to $A$, written as $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A$.

The problem states a crucial geometric constraint: the observer sees $B$ moving perpendicular to $A$'s line of motion. This means the relative velocity vector $\vec{v}_{BA}$ is perpendicular to $\vec{v}_A$.

### Resolving the Vectors

To make the mathematics elegant, let's align our coordinate system with airplane $A$'s path. We can resolve airplane $B$'s velocity into two components: one parallel to $A$'s motion, and one perpendicular to it.

Since $B$ is flying at $60^\circ$ and $A$ is flying at $30^\circ$, the angle between their flight paths is simply $60^\circ - 30^\circ = 30^\circ$.

Therefore, the component of $B$'s velocity parallel to $A$ is $v_B \cos 30^\circ$, and the component perpendicular to $A$ is $v_B \sin 30^\circ$.

### The Master Equation

If airplane $B$ appears to move *only* perpendicularly in $A$'s frame, it means $B$ is neither pulling ahead nor falling behind $A$ along the parallel direction. For this to be true, the parallel component of $B$'s velocity must exactly match $A$'s velocity.

This gives us our master equation:
$$v_B \cos 30^\circ = v_A$$

We know $v_A = 100\sqrt{3} 	ext{ m/s}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$. Substituting these values:
$$v_B \left(\frac{\sqrt{3}}{2}ight) = 100\sqrt{3}$$

The $\sqrt{3}$ beautifully cancels out from both sides, leaving us with:
$$v_B = 200 	ext{ m/s}$$

### Calculating the Relative Speed

Now that we know the actual speed of airplane $B$, we can find out how fast it is approaching $A$ in our relative frame. Since the parallel components cancelled out, the relative speed is entirely determined by the perpendicular component:
$$v_{BA} = v_B \sin 30^\circ$$

Substituting $v_B = 200 	ext{ m/s}$ and $\sin 30^\circ = \frac{1}{2}$:
$$v_{BA} = 200 	imes \frac{1}{2} = 100 	ext{ m/s}$$

### The Final Countdown

We are in the endgame now. In our relative frame, airplane $B$ is moving directly towards $A$'s path at a constant speed of $100 	ext{ m/s}$. The initial distance between them is $500 	ext{ m}$.

The time it takes for $B$ to cross $A$'s path is simply the distance divided by the relative speed:
$$t_0 = \frac{d}{v_{BA}}$$
$$t_0 = \frac{500}{100} = 5 	ext{ s}$$

And there we have it! In exactly **$5 	ext{ s}$**, airplane $B$ will cross $A$'s path. By shifting our frame of reference, we transformed a complex two-dimensional chase into a beautifully simple one-dimensional calculation.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e F l i g h tP a t h s$

$T h e C o n ce pt o f R e l a t i v e V e l oc i t y$

$T h e P er p e n d i c u l a r C o n d i t i o n$

$R eso l v in g V e l oc i t y o f B$

$E q u a t in g P a r a l l e l C o m p o n e n t s$

$C a l c u l a t in g S p ee d o f B$

$S o l v in g f or v_{B}$

$M a g ni t u d eo f R e l a t i v e V e l oc i t y$

$C a l c u l a t in g v_{B A}$

$T im e t o C o l l i s i o n S e t u p$

$F ina l C a l c u l a t i o n$

$C o n c l u s i o n$

The Sigma Insight: Relative Velocity

The Setup

The Power of Relative Velocity

Resolving the Vectors

The Master Equation

Calculating the Relative Speed

The Final Countdown

Similar Questions

Comprehension Passage

When and for how long can the particle C fulfil the given condition?

What is speed of the particle C at the instant $t = 5 s$ ?

What is modulus of acceleration of the particle C at the instant $t = 5 s$ ?

At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

Ship A is sailing towards north-east with velocity $v = 30 \hat{i} + 50 \hat{j}$ km/h, where $\hat{i}$ points east and $\hat{j}$ north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/h. A will be at minimum distance from B in

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in $ms^{- 1}$ ) at time $t = \frac{π}{3} s$ .

List-I

List-II

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VisualizingtheFlightPaths

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheConceptofRelativeVelocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ThePerpendicularCondition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ResolvingVelocityofB

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EquatingParallelComponents

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CalculatingSpeedofB

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SolvingforvB​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MagnitudeofRelativeVelocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CalculatingvBA​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TimetoCollisionSetup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalCalculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion