Instantaneous Velocity and Speed
The speedometer of a car does not report an average over the whole journey — it reports right now. That "right now" number is the instantaneous velocity, and defining it rigorously is the first appearance of calculus in Class 11 Physics. NCERT builds it as a limit: shrink the time interval Δt around the instant of interest and watch the average velocity Δx/Δt settle toward a definite value — the derivative dx/dt. This one idea unlocks the entire analytical side of kinematics: position functions become velocity functions by differentiation, and velocity functions become accelerations by differentiating again. NTA tests the concept through tangent-slope graph questions, "at what time is the velocity zero" reversals, and direct power-rule differentiations in integer-answer format. Work through the derivation once properly and every later chapter that uses calculus becomes easier.
What is Instantaneous Velocity? — Complete Theory
1. From average to instantaneous: the limit definition
The average velocity over [t, t + Δt] is Δx/Δt. Shrink the interval about the instant t:
v = limΔt→0 (Δx/Δt) = dx/dt — SI unit m s⁻¹; dimensions [L T⁻¹] (the limit process cannot change dimensions).
Each choice of instant t gives its own v(t) — instantaneous velocity is a function of time, whereas a single average belongs to one whole interval. Geometrically, as Δt shrinks, the chord between the two points pivots into the tangent at the point, so v(t) is the slope of the tangent to the x–t curve. In 1D the instantaneous speed is simply the magnitude: speed = |v| — a scalar with the same unit. The magnitude-only step is legitimate at an instant but not for interval averages, a contrast NTA exploits deliberately (see the trap box below).
2. Differentiation: the working tool
When position is given as a formula, instantaneous velocity is one differentiation away. The only rules this chapter needs:
| Rule | Statement | Example |
|---|---|---|
| Power rule | d(xⁿ)/dt = n xⁿ⁻¹ | d(t²)/dt = 2t; d(t³)/dt = 3t² |
| Constant multiple | d(k·f)/dt = k·df/dt | d(5t²)/dt = 10t |
| Sum rule | differentiate term by term | d(3t² + 2t)/dt = 6t + 2 |
| Constant dies | d(constant)/dt = 0 | x = 4 + t² → v = 2t (the 4 vanishes) |
Dimensional check, always: if x is in metres and t in seconds, then dx/dt must carry [L T⁻¹] — and indeed differentiating t² (dimension [T²]) against the SI second hands back a factor of seconds to the first power. If your differentiated velocity has wrong units, the position function itself was dimensionally inconsistent — go back and fix it before computing anything. This habit of dimension-checking every derived formula is inherited from the dimensions page of Chapter 1 and pays off in every later chapter.
3. Reading velocity off a position-time graph
Three slope facts convert any x–t sketch into a velocity story. First: the sign of the slope is the sign of v — rising graph, v > 0; falling graph, v < 0. Second: the steepness is the speed — a steeper tangent means faster motion, and the slope growing steeper point by point (concave-up curve) means the particle is speeding up. Third: a zero slope (horizontal tangent, often at the crest of a hump) means the particle is momentarily at rest — usually the turning point of the motion. A crest is not the end of motion: the velocity passes through zero and changes sign, which is exactly how a ball thrown upward behaves at its highest point (developed on the free fall page).
Visualising Instantaneous Velocity: the Tangent
Figure 2.3 — Tangent slope = instantaneous velocity
The solid curve is the position-time record. The dashed green line just touches the curve at the marked point P — it is the tangent there. Slide a ruler along the curve and watch the tangent's steepness change: that changing steepness is the changing instantaneous velocity.
Exam read-out: the tangent at P is steeper than the chord from the origin to P — so the instantaneous velocity at P exceeds the average velocity up to P. That single comparison is a complete JEE Main question stem. At a hump's crest the tangent goes horizontal (v = 0, momentary rest); left of the crest the slope is positive, right of it negative — the particle has reversed.
Solved Examples on Instantaneous Velocity (Step-by-Step)
Solved Example 1 — Instantaneous vs average: 14 vs 8
Q. A particle moves along the x-axis with x = 3t² + 2t (metres, seconds). Find (i) the instantaneous velocity at t = 2 s and (ii) the average velocity over [0, 2 s].
Step 1. Differentiate: v = dx/dt = d(3t²)/dt + d(2t)/dt = 6t + 2.
Step 2. At t = 2: v = 6(2) + 2 = 14 m s⁻¹ — the speedometer reading at that instant.
Step 3. Average: x(0) = 0, x(2) = 12 + 4 = 16 m → v̄ = 16/2 = 8 m s⁻¹.
Step 4. The two differ because the particle keeps speeding up (v grows from 2 to 14); the average 8 is the chord slope, the instantaneous 14 the tangent slope. Both are correct answers to different questions.
ANSWER: (i) 14 m s⁻¹; (ii) 8 m s⁻¹Solved Example 2 — Where does the particle reverse?
Q. For x = 6t − t² (SI), find the time and position at which the particle is momentarily at rest, and the velocity at t = 5 s.
Step 1. v = dx/dt = 6 − 2t. Set v = 0 → t = 3 s.
Step 2. Position at rest: x(3) = 18 − 9 = 9 m — the crest of the x–t hump.
Step 3. At t = 5 s: v = 6 − 10 = −4 m s⁻¹: moving at 4 m s⁻¹ in the −x direction, back toward the start.
ANSWER: at rest at t = 3 s, x = 9 m; v(5 s) = −4 m s⁻¹Practice Questions on Instantaneous Velocity (With Solutions)
Practice Q1 (Numerical-value type)
Q. A body's position is x = 5t² + 3 (SI). Its instantaneous velocity at t = 2 s is ______ m s⁻¹.
Solution. v = dx/dt = 10t + 0 = 10t; at t = 2 s, v = 20 m s⁻¹. The constant 3 contributes nothing — position offsets never affect velocity.
ANSWER: 20Practice Q2 (Single-correct MCQ)
Q. On a position-time graph, a straight horizontal line over an interval means the particle is: (a) moving with constant velocity (b) at rest throughout the interval (c) moving with constant acceleration (d) reversing direction continuously
Solution. Horizontal line → zero slope at every instant → v = 0 throughout → the particle stays at rest. (Constant velocity would need a straight slanted line.)
ANSWER: (b)Practice Q3 (Assertion–Reason)
Q. Assertion (A): The instantaneous speed of a particle equals the magnitude of its instantaneous velocity. Reason (R): Speed is a scalar while velocity is a vector, and in 1D the vector's direction is fully carried by its sign.
Options: (a) both true, R explains A · (b) both true, R does not explain A · (c) A true, R false · (d) A false, R true
Solution. At an instant the only difference between speed and velocity is direction; in 1D direction is just the sign, which magnitude strips away. Both statements are true and R explains A.
ANSWER: (a)Key Formulas & Takeaways
| Rule | Statement | Fast example |
|---|---|---|
| Definition | v = limΔt→0 Δx/Δt = dx/dt | x = 3t² + 2t → v = 6t + 2 |
| Instantaneous speed | |v| (magnitude only) | v = −4 m s⁻¹ → speed 4 m s⁻¹ |
| Graph read | tangent slope = v at that instant | horizontal tangent → v = 0 |
| Power rule | d(xⁿ)/dt = n xⁿ⁻¹; constants die | x = 5t² + 3 → v = 10t |
| Reversal | v = 0 at turning points | x = 6t − t² → rest at t = 3 s |
| Dimensional check | [dx/dt] = [L T⁻¹] | metres per second ✔ |
| Next step | a = dv/dt = d²x/dt² | v = 6 − 2t → a = −2 m s⁻² |
FAQs on Instantaneous Velocity and Speed
What is instantaneous velocity in simple words?
It is the velocity at one exact moment — the limit of Δx/Δt as the interval shrinks to zero, written v = dx/dt. On a position-time graph it is the slope of the tangent at that instant.
How is instantaneous speed related to instantaneous velocity?
In one dimension the instantaneous speed equals the magnitude of the instantaneous velocity, |dx/dt|. The equivalence holds instantaneously, but not for averages — average speed need not equal |average velocity|.
What does a zero slope on a position-time graph mean?
The instantaneous velocity is zero at that moment — the particle is momentarily at rest, for example at the top of a vertical throw. The acceleration there need not be zero.
How do you differentiate x = 3t² + 2t to get velocity?
Apply the power rule d(xⁿ)/dt = n xⁿ⁻¹ to each term: d(3t²)/dt = 6t and d(2t)/dt = 2, while the derivative of a constant is zero. So v = dx/dt = 6t + 2 (SI units: m s⁻¹).
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