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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Understanding the Setup

Step-by-Step Solution

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