Average velocity — displacement divided by time
The previous subtopic, Position, Path Length and Displacement, set up that displacement is the raw material of velocity. NCERT defines average velocity as displacement divided by time taken. If the body is at $x_1$ at instant $t_1$ and at $x_2$ at $t_2$, then
$$\bar v = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t}$$The bar flags it as an average. Strictly a vector, in 1D the direction collapses to a sign: positive $\bar v$ means net displacement along the chosen positive axis, negative means opposite. SI unit is m s$^{-1}$; $1$ km h$^{-1} = \tfrac{5}{18}$ m s$^{-1}$.
Three properties to internalise. $\bar v$ depends only on the endpoints of the interval. It inherits the sign of displacement, so any round trip gives $\bar v = 0$. And the magnitude of $\bar v$ is generally not the average of the two endpoint speeds — the most testable trap in the chapter.
The position of an object moving along the x-axis is $x = a + bt^2$, where $a = 8.5$ m and $b = 2.5$ m s$^{-2}$. What is the average velocity between $t = 2.0$ s and $t = 4.0$ s? (NCERT Example 2.1)
$x(4.0) = 8.5 + 2.5 \times 16 = 48.5$ m. $x(2.0) = 8.5 + 2.5 \times 4 = 18.5$ m.
$$\bar v = \frac{x(4.0) - x(2.0)}{4.0 - 2.0} = \frac{48.5 - 18.5}{2.0} = \frac{30}{2.0} = 15 \text{ m s}^{-1}$$
The constant $a$ drops out, exactly as expected — average velocity is insensitive to a shift in origin.
Average speed — path length divided by time
NCERT defines average speed as total path length divided by the total time interval.
$$\bar v_\text{speed} = \frac{\text{total path length}}{\text{total time}}$$Average speed is a scalar: no sign, never below zero, sensitive to the entire trajectory. Where average velocity asks "where did you end up?", average speed asks "how much road did you cover?" The numerator distinction is the whole story: $\Delta x$ for velocity, total path length for speed.
The trap NEET tested as recently as 2023 sits in how the two add up over multi-leg trips. When a body covers equal distances at two different speeds, the average speed is not the arithmetic mean — it is the harmonic mean.
Why is the harmonic mean smaller? The slower leg takes longer, so the trip spends more than half its time at the slower speed, pulling the time-weighted average below the midpoint. The arithmetic mean is correct only when each speed is held for the same time, not the same distance.
Why average speed is always at least |average velocity|
NCERT Summary point 2 records the inequality:
$$\bar v_\text{speed} \;\ge\; |\bar v|$$It follows in one line from path length $s \ge |\Delta x|$ (covered in the previous subtopic). Divide both sides by the positive $\Delta t$ and the inequality survives. Equality holds only when the body moves in one direction along a straight line without reversing.
The corollary is the favourite NEET hook: any round trip gives $|\bar v| = 0$ while average speed remains non-zero. NCERT's note next to Exercise 2.10: "You would not like to tell the tired man on his return home that his average speed was zero!"
Instantaneous velocity — the limit as Δt → 0
Average velocity describes a stretch. To know how fast the body is moving at one particular moment, NCERT §2.2 squeezes the interval down. Take the average velocity over a small $\Delta t$ centred at $t$ and let $\Delta t \to 0$:
$$v(t) = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$$Instantaneous velocity is the derivative of position with respect to time. NCERT Table 2.1 illustrates the limiting process for $x = 0.08\,t^3$: as $\Delta t$ shrinks from $2.0$ s to $0.010$ s centred at $t = 4.0$ s, $\Delta x / \Delta t$ converges to $3.84$ m s$^{-1}$ — the value of $dx/dt$ at $t = 4.0$ s.
If you have $x(t)$ algebraically, differentiate. For $x = a + b t^n$, $v = n b t^{n-1}$; sums differentiate term by term. Remember $v$ is a function of time — substitute the chosen instant only after differentiating.
Instantaneous speed
Instantaneous speed at instant $t$ is the magnitude of instantaneous velocity at the same instant: $|v(t)|$. NCERT Exercise 2.11 makes this point — over a zero-duration "interval" the body covers zero path length and zero displacement, so the limit ratios coincide in magnitude. So:
- Over an interval: average speed $\ge |$average velocity$|$, with strict inequality whenever the body reverses direction.
- At an instant: instantaneous speed $= |$instantaneous velocity$|$, always.
Reading instantaneous velocity off an x–t graph
NCERT Fig. 2.1 builds the argument visually. Pick a point $P$ on the x–t curve at $t = 4$ s. The chord from $P_1$ at $t = 3$ s to $P_2$ at $t = 5$ s has slope equal to the average velocity over that interval. Shrink the interval; the chord rotates. In the limit $\Delta t \to 0$, the chord becomes the tangent to the curve at $P$, and its slope is the instantaneous velocity at $t = 4$ s.
"The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant."
NCERT Class 11 Physics, §2 Summary point 2
The slope reading is signed. Tangent climbs left-to-right $\Rightarrow$ $v > 0$ (motion in $+x$). Descends $\Rightarrow$ $v < 0$. Horizontal $\Rightarrow$ $v = 0$ (momentarily at rest). Steepness encodes magnitude.
Uniform vs non-uniform motion
In uniform motion, velocity is constant. The x–t graph is a straight line; chord-slope and tangent-slope match everywhere. NCERT §2.2: "for uniform motion, velocity is the same as the average velocity at all instants". In non-uniform motion, the x–t curve has varying slope. Average velocity depends on which interval you pick; instantaneous velocity depends on which instant you query. They coincide only when the averaging interval shrinks to a point — which is the definition of instantaneous velocity itself.
Comparison table — the four quantities at a glance
Keep four ideas distinct: average velocity, instantaneous velocity, average speed, instantaneous speed.
| Quantity | Definition | Vector or scalar | Can be negative? | Graph reading |
|---|---|---|---|---|
| Average velocity $\bar v$ | $\dfrac{\Delta x}{\Delta t}$ (displacement ÷ time) | Vector (signed scalar in 1D) | Yes | Slope of chord on x–t |
| Instantaneous velocity $v$ | $\dfrac{dx}{dt}$ (limit as $\Delta t \to 0$) | Vector (signed scalar in 1D) | Yes | Slope of tangent on x–t |
| Average speed | $\dfrac{\text{path length}}{\Delta t}$ | Scalar | No ($\ge 0$) | Total distance plotted / time |
| Instantaneous speed | $|v|$ at instant $t$ | Scalar | No ($\ge 0$) | Magnitude of tangent slope |
The two columns that decide answers are "Can be negative?" and "Graph reading". Velocity carries a sign and reads off slope; speed is non-negative and reads off the magnitude of slope or cumulative path length.
Two worked NEET-style examples
A motorcyclist covers the first half of a $120$ km journey at $40$ km h$^{-1}$ and the second half at $60$ km h$^{-1}$. Find the average speed. (Adapted from NIOS Terminal Exercise 4.)
Plan. Total distance ÷ total time. Do not average speeds directly.
$D = 60 + 60 = 120$ km. $t_1 = 60/40 = 1.5$ h, $t_2 = 60/60 = 1.0$ h. $T = 2.5$ h.
$$\bar v_\text{speed} = \frac{D}{T} = \frac{120}{2.5} = 48 \text{ km h}^{-1}$$
Answer: $48$ km h$^{-1}$. Harmonic-mean check: $\dfrac{2 \cdot 40 \cdot 60}{100} = 48$. The arithmetic-mean answer of $50$ is wrong by $2$ km h$^{-1}$ — small in absolute terms but enough to flip an option on a NEET paper.
A particle moves along the x-axis with position $x = 3 t^2 + 2 t$ (m, s). Find (a) instantaneous velocity at $t = 2$ s; (b) average velocity over $[0, 2]$ s; (c) instantaneous speed at $t = -1$ s.
Plan. Differentiate $x(t)$; substitute. For (b) use $\Delta x / \Delta t$.
$$v(t) = \frac{dx}{dt} = 6t + 2 \quad (\text{m s}^{-1})$$
(a) $v(2) = 14$ m s$^{-1}$. (b) $x(0) = 0$, $x(2) = 16$ m, so $\bar v = 16/2 = 8$ m s$^{-1}$. (Cross-check: for constant-acceleration motion, $\bar v = \tfrac{v(0) + v(2)}{2} = \tfrac{2 + 14}{2} = 8$.) (c) $v(-1) = -4$ m s$^{-1}$; negative sign means motion in $-x$. Instantaneous speed $= |v(-1)| = 4$ m s$^{-1}$.
Answers: 14, 8, 4 m s$^{-1}$. Note $v(2) \ne \bar v_{[0,2]}$ because the motion has non-zero (constant) acceleration $a = 6$ m s$^{-2}$.
What this subtopic locked in
- Average velocity = $\Delta x / \Delta t$. Signed in 1D. Depends only on the endpoints of the interval.
- Average speed = total path length / total time. Scalar, never negative. Depends on the whole trajectory.
- Harmonic-mean trap: equal-distance two-speed trip $\Rightarrow$ average speed $= \dfrac{2 v_1 v_2}{v_1 + v_2}$, below the arithmetic mean.
- Inequality: average speed $\ge |$average velocity$|$; equality only for unreversed straight-line motion. For round trips, average velocity is zero.
- Instantaneous velocity $= \lim_{\Delta t \to 0} \Delta x / \Delta t = dx/dt$. Slope of the tangent on the x–t graph.
- Instantaneous speed $= |v(t)|$. At a single instant, no distinction from instantaneous velocity except sign.
- Uniform motion: x–t is a straight line; average velocity over any interval = instantaneous velocity at every instant. Non-uniform motion: x–t is curved; the two diverge.