A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car h

NCERT Exemplar · Q19

Q. A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of 18 m/h while the other has the speed of 27 km/h. The bird starts moving from first car towards the other and is moving with the speed of 36 km/h and when the two cars were separted by 36 km. What is the total distance covered by the bird? What is the total displacement of the bird?

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Concept understanding — Displacement Magnitude

Displacement Magnitude — The Straight-Line Shortcut

Imagine you walk 3 steps east, then 4 steps north. You end up at a spot that's not 7 steps away from where you started — it's only 5 steps away, diagonally. That 5 steps is your displacement magnitude.

Here's the core idea: displacement magnitude is the straight-line distance between where you began and where you ended. It doesn't care about the twists and turns of your actual path. It's the "as the crow flies" distance.

The Precise Definition

Displacement is a vector — it has both a direction and a magnitude. The magnitude of displacement (often written as $|\vec{s}|$ or simply $s$ ) is the length of that vector. Mathematically, if your initial position is $(x_1, y_1)$ and your final position is $(x_2, y_2)$ , then:

|\vec{s}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

This is just the distance formula from coordinate geometry. For the 3-step east, 4-step north example:

|\vec{s}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ units}

Watch out

Never confuse displacement magnitude with total distance travelled. In the example above, the total distance walked was $3 + 4 = 7$ units, but the displacement magnitude was only 5 units. They are equal only when you move in a perfectly straight line without changing direction.

Why This Matters in Physics

In kinematics problems, displacement magnitude tells you the net effect of motion. When a car drives around a circular track and returns to the starting point, its displacement magnitude is zero — even though it travelled hundreds of metres. The car ended up exactly where it began.

For motion along a straight line (say, the x-axis), the displacement magnitude simplifies to:

|\vec{s}| = |x_2 - x_1|

That's just the absolute difference between final and initial positions. No square roots needed.

A Quick Check

If a particle moves from $x = 2$ m to $x = -3$ m, what's the displacement magnitude? …

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The bird flies continuously until the cars meet. Its total distance equals speed × time-to-collision = 28.8 km. Its displacement is the straight-line separation between start and end points = 14.4 km.

Understanding the Setup

The bird doesn't make a fixed number of trips; it flies back and forth continuously until the two cars collide. The key insight is that the bird is in the air for exactly as long as it takes the cars to meet. During that entire time, the bird covers ground at its constant speed.

Distance measures the entire path length traced out—every zig and zag. Displacement measures only the straight-line separation between where the bird started and where it ended.

Step-by-Step Solution

1. Convert all speeds to consistent units

The first car's speed is given as 18 m/h, which appears to be a typo for 18 km/h (since the other speeds are in km/h). Taking it as 18 km/h:

v_1 = 18 \text{ km/h}, \quad v_2 = 27 \text{ km/h}, \quad v_{\text{bird}} = 36 \text{ km/h}

Initial separation: $d_0 = 36$ km.

2. Find the time until the cars meet

The cars approach each other, so their relative speed is the sum:

v_{\text{rel}} = v_1 + v_2 = 18 + 27 = 45 \text{ km/h}

Time to close the 36 km gap:

t = \frac{d_0}{v_{\text{rel}}} = \frac{36}{45} = 0.8 \text{ hours}

3. Calculate the total distance covered by the bird

The bird flies at 36 km/h for the entire 0.8 hours, regardless of how many times it reverses direction:

\text{Distance} = v_{\text{bird}} \times t = 36 \times 0.8 = 28.8 \text{ km}

Tip

You don't need to track individual legs of the bird's journey. The bird is always moving at constant speed, so distance = speed × total time in flight.

4. Determine the bird's displacement

The bird starts at the first car (position $x = 0$ at $t = 0$ ). When the cars meet at $t = 0.8$ h, the first car has traveled: …

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Concept: Verifying the Shortcut by Actually Summing the Bird's Individual Legs

Method: The Telescoping-Partition Argument — Compute a Few Real Legs, Then Show Why Their Infinite Sum MUST Equal the Total Time

The stored answer's own [!TIP] says "you don't need to track individual legs" — this method does exactly that, on purpose, as a genuine cross-check: it computes the first two of the bird's real legs explicitly, observes the pattern, and then proves (without summing an infinite series numerically) why the total must still come out to the same $28.8\ \text{km}$ .

Steps

Convert to consistent units and set up the closing speed of the cars, exactly as before: $v_1=18$ , $v_2=27$ , $v_{\text{bird}}=36\ \text{km/h}$ ; cars close their $36\ \text{km}$ gap at $v_1+v_2=45\ \text{km/h}$ , meeting after total time $T=36/45=0.8\ \text{h}$ .
Leg 1 — bird flies from car 1 toward car 2. The bird and car 2 approach each other (bird toward car 2, car 2 toward the bird) at combined closing speed $v_{\text{bird}}+v_2 = 36+27=63\ \text{km/h}$ . Time for this leg, until the bird meets car 2:

t_1 = \frac{36}{63} = \frac{4}{7}\ \text{h} \approx 0.5714\ \text{h}

Distance the bird covers in this leg: $d_1 = 36\times\dfrac{4}{7} = \dfrac{144}{7}\approx20.57\ \text{km}$ .

Find the new gap for leg 2. In time $t_1$ , the two cars have closed $45\times\tfrac47 = \tfrac{180}{7}\ \text{km}$ of their original $36\ \text{km}$ gap, leaving

\text{gap after leg 1} = 36 - \frac{180}{7} = \frac{72}{7}\approx10.286\ \text{km}

Leg 2 — bird flies back from car 2 toward car 1. Now closing speed is $v_{\text{bird}}+v_1=36+18=54\ \text{km/h}$ :

t_2 = \frac{72/7}{54} = \frac{4}{21}\ \text{h}\approx0.1905\ \text{h}, \qquad d_2 = 36\times\frac{4}{21}=\frac{48}{7}\approx6.857\ \text{km}

Notice the partial sum is already converging toward $T=0.8\ \text{h}$ : $t_1+t_2 = \tfrac47+\tfrac4{21} = \tfrac{12}{21}+\tfrac{4}{21}=\tfrac{16}{21}\approx0.762\ \text{h}$ — already within $0.038\ \text{h}$ of the total $0.8\ \text{h}$ after just two legs, with infinitely many (ever-shrinking) legs still to come.
The key structural argument — why the infinite sum is EXACTLY $T$ , without computing every term. Every leg is bounded by two consecutive meetings: the bird with one car, then the bird with the other. These legs partition the entire time interval $[0,T]$ with no gaps and no overlaps — at literally every instant before the cars collide, the bird is airborne on exactly one leg or another. A sum of durations that exactly, exhaustively partitions an interval of length $T$ must sum to precisely $T$ — this is true regardless of how many legs there are or how quickly they shrink, by the very definition of a partition, not by evaluating a geometric series formula.

\sum_{i=1}^{\infty} t_i = T = 0.8\ \text{h} \quad(\text{guaranteed by the partition, confirmed numerically by the fast-converging partial sum in Step 5})

Total distance, now from the actual sum of individual legs (not the shortcut): $\text{Total distance} = \sum_i d_i = \sum_i v_{\text{bird}}\,t_i = v_{\text{bird}}\sum_i t_i = 36\times0.8 = 28.8\ \text{km}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$\theta$
›Reveal solutionSolution
The key idea is that at maximum height, each projectile has zero vertical velocity and has covered half its horizontal range; the horizontal separation between them is simply the sum of their individual half‑ranges, which simplifies to $\frac{u^2 \sin 2\theta}{2g}$ .

Concept & Intuition

When two projectiles are launched from the same point with the same speed but at complementary angles ( $\theta$ and $90^\circ - \theta$ ), their trajectories are symmetric. At the instant each reaches its maximum height, the vertical velocity is zero, and the horizontal velocity remains constant ( $u\cos\theta$ for the first, $u\sin\theta$ for the second). The horizontal distance each has traveled from the launch point at that moment is exactly half its total range. The total separation between them is the sum of these two half‑ranges, because they are thrown in opposite directions.

Time to reach maximum height For a projectile launched at angle $\theta$ with speed $u$ , the vertical component is $u\sin\theta$ . At the top, $v_y = 0$ , so

$0 = u\sin\theta - g t \quad\Rightarrow\quad t = \frac{u\sin\theta}{g}.$

Horizontal distance covered in that time Horizontal velocity is constant: $u\cos\theta$ . So the horizontal distance from the launch point at maximum height is

$x_1 = (u\cos\theta) \cdot \frac{u\sin\theta}{g} = \frac{u^2 \sin\theta \cos\theta}{g}.$

For the second projectile (angle $90^\circ - \theta$ ) Its horizontal velocity is $u\cos(90^\circ - \theta) = u\sin\theta$ , and its time to maximum height is

$t' = \frac{u\sin(90^\circ - \theta)}{g} = \frac{u\cos\theta}{g}.$
Hence its horizontal distance from the launch point at maximum height is
$x_2 = (u\sin\theta) \cdot \frac{u\cos\theta}{g} = \frac{u^2 \sin\theta \cos\theta}{g}.$

Separation when both are at their maximum heights Since they are thrown in opposite directions, the total horizontal distance between them is

$x_1 + x_2 = \frac{u^2 \sin\theta \cos\theta}{g} + \frac{u^2 \sin\theta \cos\theta}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}.$
Using the identity $2\sin\theta\cos\theta = \sin 2\theta$ , we get
$\text{Separation} = \frac{u^2 \sin 2\theta}{g}.$
Watch out

A common mistake is to think the separation is the difference of the half‑ranges, but because the projectiles go in opposite directions, the distances add. …
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TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$\dfrac{\pi}{12}$
›Reveal solutionSolution
The maximum range occurs at $45°$ and gives us the launch speed; projecting at $15°$ yields half that maximum range because $\sin(30°) = \frac{1}{2}$ .

The range of a projectile depends on both the launch speed and the angle. Maximum range is achieved at $45°$ (or $\frac{\pi}{4}$ radians), and knowing this maximum lets us work backward to find the launch speed. Once we have the speed, we can calculate the range at any other angle.

The range formula is
$R = \frac{u^2 \sin(2\theta)}{g}$
where $u$ is the launch speed, $\theta$ is the angle of projection, and $g$ is acceleration due to gravity.

At maximum range, $\theta = \frac{\pi}{4}$ , so $\sin(2\theta) = \sin\left(\frac{\pi}{2}\right) = 1$ . This gives
$R_{\text{max}} = \frac{u^2}{g} = 80 \text{ m}$
This tells us that $\frac{u^2}{g} = 80$ m, which is the key quantity we need.

Now we find the range when the projectile is launched at $\theta = \frac{\pi}{12}$ (which is $15°$ ):

Calculate the angle term: We need $\sin(2\theta)$ where $\theta = \frac{\pi}{12}$ .

$2\theta = 2 \times \frac{\pi}{12} = \frac{\pi}{6}$
$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ …
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TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
$0$
›Reveal solutionSolution
When two particles in SHM with the same amplitude and frequency pass each other at half their amplitude while moving in opposite directions, their phase difference is $\boxed{2\pi/3}$ .

The motion of a particle executing Simple Harmonic Motion (SHM) can be described by its displacement from the equilibrium position as a function of time. The key to solving this problem lies in correctly interpreting the conditions given for the displacement and velocity of the two particles at the moment they pass each other.

Concept and Intuition

A particle undergoing SHM has a displacement $x(t)$ given by:
$x(t) = A \sin(\omega t + \phi)$
where $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is time, and $\phi$ is the initial phase constant. The term $(\omega t + \phi)$ is the instantaneous phase of the particle.

The velocity $v(t)$ of the particle is the time derivative of its displacement:
$v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$
The problem states that both particles have the same frequency ( $\omega$ ) and same amplitude ( $A$ ). Let their initial phase constants be $\phi_1$ and $\phi_2$ . We are looking for the phase difference, which is $|\phi_1 - \phi_2|$ .

When the particles "pass each other," it means their displacements are equal at that instant. The condition "moving in opposite direction" means their velocities must have opposite signs at that same instant. The specific displacement is given as "half their amplitude," meaning $x = A/2$ .

We will set up the equations for displacement and velocity for both particles, apply these conditions, and then solve for the phase difference.

Step-by-Step Derivation

Set up the equations for displacement and velocity: Let the displacement equations for the two particles be:

$x_1(t) = A \sin(\omega t + \phi_1)$ $x_2(t) = A \sin(\omega t + \phi_2)$
Their corresponding velocity equations are:
$v_1(t) = A\omega \cos(\omega t + \phi_1)$ $v_2(t) = A\omega \cos(\omega t + \phi_2)$
We are interested in the phase difference, $\Delta\phi = |\phi_1 - \phi_2|$.

2. Apply the displacement condition:

At the moment they pass each other, their displacements are equal and half their amplitude. Let this time be $t_0$ .
$x_1(t_0) = x_2(t_0) = \frac{A}{2}$
Substituting this into the displacement equations:
$A \sin(\omega t_0 + \phi_1) = \frac{A}{2} \implies \sin(\omega t_0 + \phi_1) = \frac{1}{2}$ $A \sin(\omega t_0 + \phi_2) = \frac{A}{2} \implies \sin(\omega t_0 + \phi_2) = \frac{1}{2}$
Let $\theta_1 = \omega t_0 + \phi_1$ and $\theta_2 = \omega t_0 + \phi_2$. Then we have:
$\sin(\theta_1) = \frac{1}{2} \quad \text{and} \quad \sin(\theta_2) = \frac{1}{2}$
This means that $\theta_1$ and $\theta_2$ can be $\pi/6$ (first quadrant) or $5\pi/6$ (second quadrant), or angles coterminal with these.

3. Apply the velocity condition:

At the same instant $t_0$ , the particles are moving in opposite directions. This means their velocities have opposite signs:
$v_1(t_0) = -v_2(t_0)$
Substituting the velocity equations:
$A\omega \cos(\omega t_0 + \phi_1) = -A\omega \cos(\omega t_0 + \phi_2)$
Dividing by $A\omega$ (since $A \neq 0, \omega \neq 0$):
$\cos(\theta_1) = -\cos(\theta_2)$

Combine conditions to find the phase difference:

We have two conditions for $\theta_1$ and $\theta_2$ :

$\sin(\theta_1) = \frac{1}{2}$ and $\sin(\theta_2) = \frac{1}{2}$

$\cos(\theta_1) = -\cos(\theta_2)$ From condition (i), $\theta_1$ and $\theta_2$ must be angles whose sine is $1/2$ . These are typically $\pi/6$ or $5\pi/6$ (within the range $[0, 2\pi)$ ). Let's consider the possible values for $\theta_1$ :

Case 1: If $\theta_1 = \pi/6$ Then $\cos(\theta_1) = \cos(\pi/6) = \frac{\sqrt{3}}{2}$ . From condition (ii), $\cos(\theta_2) = -\cos(\theta_1) = -\frac{\sqrt{3}}{2}$ . …
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Understanding the Setup

Step-by-Step Solution

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