Q.(a) Differentiate between distance and displacement.
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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 ∣s∣ or simply s) is the length of that vector. Mathematically, if your initial position is (x1,y1) and your final position is (x2,y2), then:
∣s∣=(x2−x1)2+(y2−y1)2
This is just the distance formula from coordinate geometry. For the 3-step east, 4-step north example:
∣s∣=32+42=9+16=25=5 units
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:
∣s∣=∣x2−x1∣
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? …
Distance is the total path length covered (a scalar), while displacement is the straight-line change in position (a vector). Going 2 km to school and 2 km back covers a total distance of 4 km, but the displacement on returning home is zero. …
Distance is total path length (scalar); displacement is the straight-line position change (vector). Round trip home-school-home: distance = 4 km, displacement = 0.
(a) Difference between distance and displacement:
- Distance is the total length of the actual path travelled by a body. It is a scalar quantity, always positive, and does not depend on direction.
- Displacement is the change in position, i.e. the shortest (straight-line) directed distance from the initial to the final position. It is a vector quantity and can be positive, negative or zero.
(b) The home and school are 2 km apart. In the morning you travel from home to school (2 km) and in the evening from school back to home (2 km). …
Showing the 12 most recent of 14 on this concept.
- CBSE 2026Set ANNUAL1 markQ.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 km.
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- CBSE 2025Set ANNUAL1 markMCQQ.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 11:72 when π is taken as 22/7.
Let the radius of the circle be r.
Distance travelled (arc length of a quarter circle):
d=(1/4)(2πr)=πr/2
Using π=22/7: d=(22/7)(r)/2=11r/7
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- CBSE 2024Set ANNUAL1 markMCQQ.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 = 2πr …
- CBSE 2023Set annual1 markQ.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.
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- CBSE 2023Set ANNUAL1 markMCQQ.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.
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- CBSE 2022Set TERM11 markMCQQ.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.
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- CBSE 2022Set TERM11 markMCQQ.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).
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- CBSE 2022Set ANNUAL1 markMCQQ.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 =πr (half the circumference). …
- CBSE 2022Set ANNUAL1 markQ.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. …
- CBSE 2021Set ANNUAL1 markMCQQ.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 =πr (half of the full circumference 2πr).
…
- CBSE 2021Set TERM11 markMCQQ.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.
…
- CBSE 2020Set ANN1 markQ.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) …
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