A particle's position moves from $\vec{r}_1 = 3\hat{i}+4\hat{j}$ to $\vec{r}_2 = \hat{i}+2\hat{j}$. Calculate the displacement vector $\Delta\vec{r}$ and draw $\vec{r}_1$, $\vec{r}_2$ and $\Delta\vec{r}$ in a two dimensional Cartesian coordinate system.

A particle's position moves from vecr₁ = 3hati+4hatj to vecr₂ = hati+2hatj. Calculate the displacement vector

IV. Exercises · Q2

\vec{r}_1 = 3\hat{i}+4\hat{j}

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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?

The displacement vector is $\vec{s} = (-3 - 2) = -5$ m. Its magnitude is $| -5 | = 5$ m. The negative sign only tells you the direction (leftwards), but the magnitude — the distance between the two points — is 5 metres.

Important

Displacement magnitude is always non-negative. It's a length, and lengths can't be negative. The sign of the displacement vector tells you direction; the magnitude tells you how far apart the start and end points actually are.

Looking up "Displacement Magnitude: definition, formula & real-world examples" is a good habit before an exam, and it is worth knowing that Displacement Magnitude is a core, NCERT-aligned topic from the Motion in a Straight Line portion of the Class 11 Physics curriculum, and it is tested regularly in CBSE board exams as well as in JEE Main and NEET. Cross-checking this explanation against the relevant NCERT Physics chapter and solving a few past-year questions will round out your preparation.

Step 1. Given $\vec r_1=3\hat i+4\hat j$ and $\vec r_2=\hat i+2\hat j$ .

Step 2. $\Delta\vec r=\vec r_2-\vec r_1=(1-3)\hat i+(2-4)\hat j=-2\hat i-2\hat j$ .

Step 3. Geometrically: plot $\vec r_1$ as an arrow from the origin to $(3,4)$ , $\vec r_2$ as an arrow from the origin to $(1,2)$ , and $\Delta\vec r$ as the arrow drawn directly from the tip of $\vec r_1$ , $(3,4)$ , to the tip of $\vec r_2$ , $(1,2)$ — pointing down and to the left, consistent with the components $(-2,-2)$ .

✓Final answer

\Delta\vec r=\vec r_2-\vec r_1=-2\hat i-2\hat j

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

Showing the 12 most recent of 14 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. A boy went to market from his home, the distance of market to his home is 5 km. He found that market was closed, then he returned to his home. Total displacement covered by boy is ............. .$
›Reveal solutionSolution

Displacement depends only on the start and end points; since the boy returns to his starting point, his net displacement is zero.

Displacement is a vector quantity equal to the straight-line change in position, i.e. (final position) $-$ (initial position). Distance, by contrast, is the total path length actually travelled, regardless of direction.

The boy starts at home, walks 5 km to the market (distance = 5 km), finds it closed, and walks back the same 5 km to home (distance = 5 km). So the total distance covered is $5+5=10\text{ km}$ .

But his final position is exactly the same as his initial position (he is back home), so the net displacement is:

$\vec{s} = \vec{x}_{\text{final}} - \vec{x}_{\text{initial}} = \vec{x}_{\text{home}} - \vec{x}_{\text{home}} = 0$

✓Final answer
$Total displacement = 0 (zero) — although the total distance travelled is 10 km.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. A particle moves along the curved path of a quarter circle, calculate the ratio of distance to displacement:$
(a) 11 : 7
(b) 11 : 7√2
(c) 7 : 11
(d) 7 : 11√2
›Reveal solutionSolution

Distance is the arc length along the quarter circle; displacement is the straight-line chord joining the start and end points. Their ratio works out to $11 : 7\sqrt2$ when $\pi$ is taken as $22/7$ .

Let the radius of the circle be $r$ .

Distance travelled (arc length of a quarter circle):

$d = (1/4)(2\pi r) = \pi r/2$

Using $\pi = 22/7$ : $d = (22/7)(r)/2 = 11r/7$

Displacement (straight line from start to end point of the quarter circle): the two radii at the ends of a quarter circle are perpendicular to each other, so the chord joining them is the hypotenuse of a right triangle with both legs $= r$ :

$\text{displacement} = \sqrt{r^2 + r^2} = r\sqrt2$

Ratio:

$d : \text{displacement} = (11r/7) : (r\sqrt2) = 11 : 7\sqrt2$

✓Final answer
$= 11 : 7\sqrt2$
CBSE 2024Set ANNUAL1 markMCQ
$Q. A particle completes circular path of radius r, displacement of particle will be :$
(a) 2πr
(b) 2π
(c) πr
(d) zero
›Reveal solutionSolution
Completing one full circle brings the particle back to its starting point, so displacement = 0 (even though distance travelled = 2πr).

Displacement is defined as the shortest straight-line vector joining the initial and final positions of the particle, irrespective of the path taken. Distance, by contrast, is the total length of the actual path travelled.

For a particle that completes one full revolution of a circle of radius r:

Distance travelled = circumference = $2\pi r$

Final position coincides exactly with the initial position

Since initial position = final position, the displacement vector has zero magnitude.

✓Final answer
$Displacement after one complete circular path = zero. Option (d) zero.$
CBSE 2023Set annual1 mark
$Q. A man arrived at Delhi Railway Station and wanted to go to his relative's house 10 km away from the station. He hired a taxi to reach the destination. The driver followed a long path of 25 km to reach the destination in one hour and charged for 25 km from the man. Comment on the behaviour of driver.$
›Reveal solutionSolution

The driver behaved dishonestly by taking an unnecessarily long route and overcharging the passenger for it.

The straight-line distance between the railway station and the relative's house is only 10 km, but the driver deliberately drove a longer path of 25 km and then charged the passenger the fare for the full 25 km travelled.

This behaviour is dishonest and unethical: the driver exploited the passenger's unfamiliarity with the city to take a needlessly longer route purely to increase his fare/earnings, at the direct cost of the passenger's money and time. A responsible and honest driver should take the shortest practical route to the destination and charge fairly for the actual necessary distance, not deliberately inflate it.

This situation also nicely illustrates, in physics terms, the difference between distance (the total path length actually travelled, 25 km here) and displacement (the shortest straight-line distance between start and end points, 10 km here) — the driver exploited this very gap between the two for unfair personal gain.

✓Final answer
$The driver's behaviour was dishonest/unethical: he took a needlessly longer route (25 km instead of the direct 10 km) and charged the passenger for the extra distance, exploiting the passenger for personal gain.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. An object is moving on circular path of radius R. Displacement of object in T/2 time will be:$
(a) πR
(b) 2R
(c) 2πR
(d) πR^2
›Reveal solutionSolution

Displacement in half a revolution = diameter = 2R.

In one full period T the particle completes one revolution. In time T/2 it covers half the circle and arrives at the point diametrically opposite its start.

Displacement is the straight-line distance between start and end points = the diameter = 2R. (The path length, distance, would be half the circumference = πR.)

✓Final answer
$(B) 2R.$
CBSE 2022Set TERM11 markMCQ
$Q. A runner covers a circular path of radius R in 40 seconds. His displacement after 2 minute 20 seconds is$
(1) zero
(2) 2 R
(3) 2 pi R
(4) 7 pi R
›Reveal solutionSolution

2 minutes 20 seconds = 3.5 revolutions of the 40-second lap. After a half-integer number of laps, the runner sits at the point diametrically opposite the start -- displacement = diameter = 2R.

Time for one complete circuit = 40 s.

Total time elapsed = 2 min 20 s = 140 s.

Number of laps = 140 / 40 = 3.5 laps.

3 complete laps bring the runner back to the starting point (displacement contribution = 0). The remaining 0.5 lap (half the circle) takes him to the point diametrically opposite the start. The straight-line distance between two diametrically opposite points on a circle of radius R is the diameter, 2R.

✓Final answer
$(2) 2R.$
CBSE 2022Set TERM11 markMCQ
$Q. A body starts from a point A, travels to a point B at a distance of 1.5 km and returns to A. If he takes one hour to do so, his average velocity is$
(1) 3 km/h
(2) zero
(3) 1.5 km/h
(4) 2 km/h
›Reveal solutionSolution

Average velocity depends on DISPLACEMENT (a vector), not distance. A round trip back to the start has zero net displacement, so average velocity is zero even though the body clearly moved and took time to do so.

Average velocity = (total displacement) / (total time taken).

The body travels from A to B (1.5 km) and back from B to A (1.5 km), ending exactly where it started. Its net displacement (straight-line change in position from start to end) is therefore 0 km, even though the total distance covered is 3 km.

Average velocity = 0 km / 1 h = 0 km/h.

(Note: the AVERAGE SPEED, by contrast, would be total distance/time = 3 km/1 h = 3 km/h -- option (1) is a distractor testing exactly this distinction.)

✓Final answer
$(2) zero.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. A particle completes semicircular path of radius r, displacement travelled by particle will be:$
(a) r/4
(b) r/2
(c) 2r
(d) 4r
›Reveal solutionSolution
On a semicircular path of radius r, the particle's displacement is the diameter, 2r — not the arc length πr.

Concept. Distance is the total path length covered; displacement is the vector from the initial position to the final position (straight line).

For a particle moving along a semicircular arc of radius r:

Distance travelled $= \pi r$ (half the circumference).

The initial and final points lie at opposite ends of a diameter of the circle, so the straight-line separation between them is $2r$ .

Hence displacement $= 2r$ , directed along the diameter joining the start and end points.

✓Final answer
$The displacement is 2r (the diameter), while the distance travelled is πr.$
CBSE 2022Set ANNUAL1 mark
$Q. A body covers a distance L m along a semicircular path. What is the magnitude of displacement of the body?$
›Reveal solutionSolution

For a semicircular path of arc length L, the displacement equals the diameter of the circle, which works out to 2L/π.

Step 1: Relate arc length to radius.

For a semicircle of radius r, the arc length (the "distance" travelled along the curved path) is:

L = πr

So the radius is:

r = L/π

Step 2: Identify the displacement.

The body starts at one end of the semicircular arc and ends at the diametrically opposite end. Displacement is the straight-line vector from start to finish — for a semicircle, that straight line is exactly the diameter of the circle.

Displacement = 2r

Step 3: Substitute r = L/π.

Displacement = 2 × (L/π) = 2L/π

Step 4: Interpret.

This illustrates the key distinction between distance (the actual path length, L, always ≥ displacement) and displacement (the straight-line shortcut between start and end points, here 2L/π ≈ 0.637L).

✓Final answer
$The magnitude of displacement is 2L/π metres.$
CBSE 2021Set ANNUAL1 markMCQ
$Q. A particle completes semicircular path of radius r. The ratio of distance travelled and displacements of particle will be :$
(a) π/4
(b) π/2
(c) 3π/4
(d) π
›Reveal solutionSolution

Distance travelled along a semicircle is half the circumference; displacement is just the diameter joining the start and end points.

For a particle moving along a semicircular arc of radius $r$ :

Distance travelled = arc length of a semicircle $= \pi r$ (half of the full circumference $2\pi r$ ).

Displacement = straight-line distance between the initial and final points, which for a semicircle are diametrically opposite = $2r$ .

$\text{Ratio} = \frac{\text{Distance}}{\text{Displacement}} = \frac{\pi r}{2r} = \frac{\pi}{2}$

✓Final answer
$= \pi/2$
CBSE 2021Set TERM11 markMCQ
$Q. A person starts his journary from his home at 9.00 A.M. to his office and come back to his home at 5.00 P.M. His office is 30 Km from his home, then the displacement in his motion is:$
(a) 30 Km
(b) Zero
(c) Not defined
(d) None of these
›Reveal solutionSolution

Displacement depends only on the initial and final position, not on the path taken; leaving home and returning home means the final position coincides with the initial position, so displacement = 0.

Displacement is defined as the vector joining the initial position to the final position of a body, irrespective of the actual path followed. Distance, on the other hand, is the total length of the path travelled.

Here the person starts at home (9:00 AM), travels 30 km to office, and returns to the SAME home (5:00 PM). The total distance travelled is 60 km (30 km each way), but since the initial position (home) and the final position (home) are identical, the displacement vector has zero magnitude.

✓Final answer
$The correct option is (b) Zero.$
CBSE 2020Set ANN1 mark
$Q. Four pairs of initial and final positions of a body along an x axis are given. Which pair gives a positive displacement of the body ?$
(a) -10 m, +15 m
(b) -5 m, -12 m
(c) 2 m, -5 m
(d) 2 m, 1m
›Reveal solutionSolution
Displacement = final position − initial position. Only the pair (-10 m, +15 m) gives a positive value.

For each pair (initial position x₁, final position x₂), displacement Δx = x₂ − x₁:

x₁ = −10 m, x₂ = +15 m → Δx = 15 − (−10) = +25 m (positive)

x₁ = −5 m, x₂ = −12 m → Δx = −12 − (−5) = −7 m (negative)

x₁ = 2 m, x₂ = −5 m → Δx = −5 − 2 = −7 m (negative)

x₁ = 2 m, x₂ = 1 m → Δx = 1 − 2 = −1 m (negative) Only option (a) gives a positive displacement.

✓Final answer
$(a) -10 m, +15 m (displacement = +25 m)$

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