Two ways to measure how fast an object moves — one based on path length, the other on displacement. Their difference is subtle but crucial.
Q. Why do we need the concept of average speed?
In everyday life, objects rarely move at a perfectly constant rate. A car in city traffic slows down at signals and speeds up on open road. A cricket ball changes speed as it travels through air. To describe such motion with a single, useful number, we use average speed.
The average speed of an object is the total distance covered divided by the total time taken.
$$\text{Average speed} = \frac{\text{Total distance covered}}{\text{Time interval}} \quad \cdots \text{(Eq. 4.1)}$$In symbols, if an object travels total distance $d$ in time interval $\Delta t$:
$$v_{av} = \frac{d}{\Delta t}$$SI unit: metre per second (m/s) | Type: Scalar quantity (no direction)
Always positive — because distance is always positive and time is always positive.
Q. What does average speed actually tell us?
It tells you at what constant speed you would need to travel to cover the same total distance in the same total time. Think of it as a "summary" speed for the entire journey — not necessarily the speed at any particular moment.
Average speed is not the average of the speeds at different moments. It is always calculated as total distance ÷ total time. If you travel 60 km/h for half the time and 40 km/h for the other half, your average speed is not 50 km/h — you must calculate total distance and total time separately.
Q. If average speed uses distance, what uses displacement?
Just as average speed is total distance over time, average velocity is total displacement over time. Because displacement is a vector (it carries direction), average velocity is also a vector.
The average velocity of an object is the total displacement divided by the total time taken.
$$\vec{v}_{av} = \frac{\text{Total displacement}}{\text{Time interval}} \quad \cdots \text{(Eq. 4.2b)}$$In symbols, if displacement vector is $\vec{s}$ and time interval is $\Delta t$:
$$\vec{v}_{av} = \frac{\vec{s}}{\Delta t}$$SI unit: metre per second (m/s) | Type: Vector quantity (has direction)
Can be zero — if the object returns to its starting position, displacement = 0, so average velocity = 0, even if average speed is not zero.
Can be negative — if net displacement is opposite to the chosen positive direction.
Q. What is the key difference between average speed and average velocity in one line?
Average speed uses the total distance (entire path length), while average velocity uses only the net displacement (straight-line shift from start to end).
Both speed and velocity have the SI unit m/s, but they measure different things. Speed answers "How much ground did you cover per second?" while velocity answers "How far did you actually shift from where you started, per second?" A marathon runner who completes a circular course covers 42 km in distance, yet ends up at the starting line — speed ≠ 0, velocity = 0.
This is the key worked example from NCERT Chapter 4 that brings both concepts to life in one situation.
Sarang swims in a 50-metre-long pool. He starts at one end (let us call it point A) and swims to the other end (point B), covering 50 m. He then turns around and swims back to point A. The total time taken for this round trip is 100 seconds.
Step 1: Identify what is given
Step 2: Calculate total distance
Total distance = distance from A to B + distance from B to A
$$d = 50 \text{ m} + 50 \text{ m} = 100 \text{ m}$$Step 3: Calculate total displacement
Sarang starts at A and ends back at A. Net change in position = 0.
$$\vec{s} = 0 \text{ m}$$Step 4: Calculate average speed
$$\text{Average speed} = \frac{\text{Total distance}}{\text{Time}} = \frac{100 \text{ m}}{100 \text{ s}} = 1 \text{ m/s}$$Step 5: Calculate average velocity
$$\vec{v}_{av} = \frac{\text{Displacement}}{\text{Time}} = \frac{0 \text{ m}}{100 \text{ s}} = 0 \text{ m/s}$$If Sarang swims only from A to B (50 m, 50 s):
In this case, average speed equals the magnitude of average velocity. This happens whenever motion is in one direction only (no back-tracking). See Section 5 for the general rule.
Q. How do we classify motion based on how speed changes over time?
When we observe objects moving around us, some move at a steady pace while others speed up, slow down, or change direction. This leads us to two fundamental categories of motion.
An object is said to be in uniform motion if it travels equal distances in equal intervals of time, no matter how small those time intervals are. The speed stays constant throughout.
In uniform motion: average speed = speed at any instant (instantaneous speed). There is no change in speed, so the average and the instantaneous values are identical.
An object is in non-uniform motion if it travels unequal distances in equal intervals of time. Its speed (or direction, or both) changes as it moves.
For non-uniform motion, the concept of average speed becomes essential because the speed at each instant is different.
| Parameter | Uniform Motion | Non-Uniform Motion |
|---|---|---|
| Distance in equal time | Equal distances | Unequal distances |
| Speed | Constant throughout | Varies with time |
| Acceleration | Zero | Non-zero |
| Position-time graph | Straight line (constant slope) | Curved line (changing slope) |
| Average speed vs instantaneous speed | Always equal | May differ |
| Common examples | Car at cruise control, light in vacuum | Ball thrown up, bus starting from stop |
Q. Is average speed always different from the magnitude of average velocity?
Not always. There is a special condition when they are equal.
The magnitude of average velocity equals average speed only when the object moves along a straight line in one direction (no reversal, no detour).
In such a case, total distance = magnitude of displacement, so:
$$|\vec{v}_{av}| = \text{average speed} \quad \text{(only for straight-line, one-direction motion)}$$In all other cases (curved paths, back-and-forth motion), the distance is greater than the magnitude of displacement, so:
$$\text{Average speed} \geq |\vec{v}_{av}|$$The equal sign holds only for straight-line, one-direction motion as stated above.
| Situation | Distance vs Displacement | Speed vs |Velocity| |
|---|---|---|
| Straight line, one direction | Distance = |Displacement| | Speed = |Velocity| |
| Back-and-forth (round trip) | Distance > |Displacement| | Speed > |Velocity| |
| Curved path | Distance > |Displacement| | Speed > |Velocity| |
| Complete circle (returns to start) | Distance = circumference, Displacement = 0 | Speed > 0, |Velocity| = 0 |
Q. What are all the important differences between speed and velocity?
| Parameter | Speed | Velocity |
|---|---|---|
| Definition | Distance covered per unit time | Displacement per unit time |
| Formula | $v = d / t$ | $\vec{v} = \vec{s} / t$ |
| Type of quantity | Scalar (magnitude only) | Vector (magnitude + direction) |
| Direction required? | No | Yes |
| Can it be zero? | Only if object is at rest | Yes — even for moving object (round trip) |
| Can it be negative? | No (always ≥ 0) | Yes (opposite to chosen positive direction) |
| SI unit | m/s | m/s |
Q. Why do we need to convert between km/h and m/s?
Speed is often given in km/h in daily life (speedometers, news reports, weather), but physics calculations require SI units (m/s). Knowing how to convert quickly is an essential skill.
1 kilometre = 1000 metres and 1 hour = 3600 seconds
$$1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}$$So to convert km/h → m/s: multiply by $\dfrac{5}{18}$
To convert m/s → km/h: multiply by $\dfrac{18}{5}$
| Speed in km/h | Calculation | Speed in m/s |
|---|---|---|
| 36 km/h | $36 \times \dfrac{5}{18} = \dfrac{180}{18}$ | 10 m/s |
| 54 km/h | $54 \times \dfrac{5}{18} = \dfrac{270}{18}$ | 15 m/s |
| 72 km/h | $72 \times \dfrac{5}{18} = \dfrac{360}{18}$ | 20 m/s |
| 108 km/h | $108 \times \dfrac{5}{18} = \dfrac{540}{18}$ | 30 m/s |
Remember: 36 km/h = 10 m/s. This is the "anchor" conversion. From here:
These values appear very frequently in physics problems, so memorising the anchor saves time.
Long before the European scientific revolution, Indian scholars were already grappling with precise ideas about position, speed, and time. The NCERT textbook highlights two remarkable traditions.
In his landmark treatise Aryabhatiya, the astronomer-mathematician Aryabhata described the motion of celestial bodies — the Sun, Moon, and planets — in terms of their speeds along circular paths. He calculated the number of rotations and revolutions in a Mahayuga (a vast cosmic time cycle), and from this derived precise average angular speeds. This was arguably the first systematic use of the concept of average velocity in an astronomical context.
Aryabhata also correctly argued that the Earth rotates on its own axis — an insight that required a clear understanding of relative motion: what appears as the motion of the stars is actually due to Earth's spin.
The mathematician Narayana Pandita, in his Ganitakaumudi, posed and solved problems that involved objects (postmen, messengers) moving between two towns at different speeds, meeting each other on the way. These are essentially problems of relative speed and average velocity — the same concepts studied in Chapter 4 today. His work shows that the mathematical treatment of motion as a computational problem has deep roots in Indian mathematics.
Why this matters: Science is a global endeavour built on contributions from many cultures. Recognising India's ancient achievements in quantitative reasoning about motion enriches our understanding of how physics developed over centuries.
Problem: A car travels 200 km due north in 3 hours. It then turns around and travels 200 km due south in 2 hours. Calculate (a) the average speed for the entire journey, and (b) the average velocity for the entire journey.
Given:
Step 1: Total distance
$$d = 200 \text{ km} + 200 \text{ km} = 400 \text{ km}$$Step 2: Total time
$$t = 3 \text{ h} + 2 \text{ h} = 5 \text{ h}$$Step 3: Net displacement
The car started at, say, point O. After going 200 km north it is at point A. Then it comes 200 km back south — it is at O again.
$$\vec{s} = +200 \text{ km} + (-200 \text{ km}) = 0 \text{ km}$$(a) Average speed:
$$\text{Average speed} = \frac{400 \text{ km}}{5 \text{ h}} = \mathbf{80 \text{ km/h}}$$(b) Average velocity:
$$\vec{v}_{av} = \frac{0 \text{ km}}{5 \text{ h}} = \mathbf{0 \text{ km/h}}$$Conclusion: The car was moving for 5 hours and covered 400 km, yet its average velocity is zero because it returned to the starting point. This is the same principle as the Sarang swimming pool example.
Problem: Priya cycles from her home to school (6 km) in 20 minutes. She then continues straight to the library, which is 4 km further in the same direction, taking 10 minutes. Calculate her average speed and average velocity for the full trip.
Given:
Step 1: Total distance = 6 + 4 = 10 km
Step 2: Total time = 20 + 10 = 30 min = 0.5 h
Step 3: Displacement — since both segments are in the same direction (straight road), displacement = 10 km (from home to library).
Average speed:
$$= \frac{10 \text{ km}}{0.5 \text{ h}} = \mathbf{20 \text{ km/h}}$$Average velocity:
$$= \frac{10 \text{ km (towards library)}}{0.5 \text{ h}} = \mathbf{20 \text{ km/h towards library}}$$Here, speed equals the magnitude of velocity because motion is along a straight line in one direction.
Problem: A train is moving at 90 km/h. Convert this speed to m/s. Also, how far (in metres) does it travel in 4 seconds at this speed?
Part (a): Convert 90 km/h to m/s
$$90 \text{ km/h} = 90 \times \frac{5}{18} \text{ m/s} = \frac{450}{18} \text{ m/s} = \mathbf{25 \text{ m/s}}$$Part (b): Distance in 4 seconds
Using distance = speed × time:
$$d = 25 \text{ m/s} \times 4 \text{ s} = \mathbf{100 \text{ m}}$$The train travels 100 m in 4 seconds.
Q1. A bus covers 120 km in 3 hours. What is its average speed in (a) km/h and (b) m/s?
(a) Average speed = 120 km ÷ 3 h = 40 km/h
(b) $40 \times \dfrac{5}{18} = \dfrac{200}{18} \approx$ 11.1 m/s
Q2. Can a body have non-zero speed but zero average velocity? Give an example.
Yes. If an object returns to its starting point, displacement = 0, so average velocity = 0, even though it covered distance and had non-zero average speed.
Example: Sarang swimming from A to B and back to A. Average speed = 1 m/s; average velocity = 0.
Q3. A car's speedometer shows 80 km/h. Is this average speed or instantaneous speed? Explain.
It is instantaneous speed — the speed at that particular moment. The speedometer responds to the current state of the engine and wheels, not the average over the whole journey. Average speed would require knowing total distance and total time for the entire trip.
Q4. What is the difference between uniform motion and non-uniform motion? Give two examples of each.
Uniform motion: Equal distances in equal time intervals. Speed is constant. Examples: (1) A fan blade rotating at fixed RPM; (2) A car on cruise control on a straight road.
Non-uniform motion: Unequal distances in equal time intervals. Speed varies. Examples: (1) A bus accelerating from a stop; (2) A ball thrown vertically upward.
Q5. Convert the following speeds to m/s: (a) 18 km/h, (b) 45 km/h, (c) 120 km/h.
(a) $18 \times \dfrac{5}{18} = \mathbf{5 \text{ m/s}}$
(b) $45 \times \dfrac{5}{18} = \dfrac{225}{18} = \mathbf{12.5 \text{ m/s}}$
(c) $120 \times \dfrac{5}{18} = \dfrac{600}{18} = \mathbf{33.33 \text{ m/s}}$
N1. Riya runs around a rectangular field 80 m × 60 m. She starts at one corner and completes one full lap in 140 s. Calculate (a) the total distance covered, (b) her displacement at the end of one lap, (c) average speed, and (d) average velocity.
(a) Perimeter = 2(80 + 60) = 2 × 140 = 280 m
(b) After one complete lap, she returns to the starting corner. Displacement = 0 m
(c) Average speed = 280 m ÷ 140 s = 2 m/s
(d) Average velocity = 0 m ÷ 140 s = 0 m/s
N2. A train travels from station A to station B, a distance of 360 km, in 4 hours. It then travels from B to station C, 240 km further in the same direction, in 3 hours. Find the average speed for the whole journey from A to C.
Total distance = 360 + 240 = 600 km
Total time = 4 + 3 = 7 h
$$\text{Average speed} = \frac{600}{7} \approx \mathbf{85.7 \text{ km/h}}$$Note: This is NOT the average of 90 km/h and 80 km/h (which would give 85 km/h). The correct method always uses total distance ÷ total time.
N3. An object moves along a straight line. It covers 30 m in the first 5 s, then remains stationary for 3 s, then covers 20 m in the opposite direction in 4 s. Find: (a) average speed, (b) average velocity (take the original direction as positive).
Total distance = 30 + 0 + 20 = 50 m
Total time = 5 + 3 + 4 = 12 s
Net displacement = +30 − 20 = +10 m (in original direction)
(a) Average speed = 50 ÷ 12 ≈ 4.17 m/s
(b) Average velocity = +10 ÷ 12 ≈ +0.83 m/s (in original direction)
N4. A motorcycle travels the first half of a 100 km journey at 50 km/h and the second half at 100 km/h. What is the average speed for the entire journey? (Hint: it is NOT 75 km/h.)
Time for first 50 km at 50 km/h = 50 ÷ 50 = 1 h
Time for second 50 km at 100 km/h = 50 ÷ 100 = 0.5 h
Total distance = 100 km; Total time = 1.5 h
$$\text{Average speed} = \frac{100}{1.5} = \mathbf{66.7 \text{ km/h}}$$This is the harmonic mean of the two speeds (for equal distances), not the arithmetic mean. The motorcycle spends more time at the slower speed, pulling the average down.
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