Is it possible for a body to have zero velocity at some instant and yet be accelerating at that very instant?

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Q. Is it possible for a body to have zero velocity at some instant and yet be accelerating at that very instant? Justify your answer with a suitable example.

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Concept understanding — Instantaneous Velocity

Instantaneous Velocity: From "How Fast" to "How Fast Right Now"

You already know average velocity. If a car travels 120 km in 2 hours, its average velocity is 60 km/h. That tells you the overall rate, but it hides everything that happened in between — the traffic jams, the sudden bursts of speed, the moments the car was completely stopped.

Now imagine you want to know the car's velocity at exactly 10:15 AM, not averaged over an hour or a minute. That's instantaneous velocity — the velocity at a single instant of time.

The Intuition: Zooming In

Think of a speedometer needle. When you drive, the needle doesn't stay fixed at 60 km/h. It jumps up when you accelerate, drops when you brake. At any given moment, the needle points to a specific number. That number is your instantaneous speed (velocity, if direction matters).

But here's the puzzle: at a single instant, the car hasn't moved any distance. How can you have a speed if $\Delta t = 0$ ? You can't divide by zero.

The trick is to shrink the time interval smaller and smaller, and see what the average velocity approaches.

The Precise Definition

Let $s(t)$ be the position of an object at time $t$ . The average velocity over a time interval $[t, t+h]$ is:

v_{\text{avg}} = \frac{s(t+h) - s(t)}{h}

Now, let $h$ get closer and closer to 0 (but never equal to 0). If the average velocity settles down to a single number as $h \to 0$ , that number is the instantaneous velocity at time $t$ :

v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}

v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}

This limit is exactly the derivative of position with respect to time. In calculus notation: $v(t) = s'(t)$ .

A Concrete Example

Suppose a ball is dropped from rest, and its height (in meters) after $t$ seconds is $s(t) = 4.9t^2$ (ignoring air resistance).

Average velocity from $t=2$ to $t=2.1$ seconds:

v_{\text{avg}} = \frac{4.9(2.1)^2 - 4.9(2)^2}{0.1} = \frac{4.9(4.41 - 4)}{0.1} = \frac{4.9 \times 0.41}{0.1} = 20.09 \text{ m/s}

Average velocity from $t=2$ to $t=2.01$ :

v_{\text{avg}} = \frac{4.9(2.01)^2 - 4.9(2)^2}{0.01} = \frac{4.9(4.0401 - 4)}{0.01} = 19.649 \text{ m/s}

Average velocity from $t=2$ to $t=2.001$ :

v_{\text{avg}} = \frac{4.9(2.001)^2 - 4.9(2)^2}{0.001} = 19.6049 \text{ m/s}

The numbers are converging to 19.6 m/s. That's the instantaneous velocity at $t=2$ seconds.

Using the derivative: $v(t) = 9.8t$ , so $v(2) = 19.6$ m/s. Matches perfectly.

Key Takeaways for Exams

Concept	Meaning	Formula
Average velocity	Total displacement ÷ total time	$\frac{\Delta s}{\Delta t}$
Instantaneous velocity	Velocity at a single moment	$\lim_{h \to 0} \frac{s(t+h)-s(t)}{h}$

Watch out

Do not confuse instantaneous velocity with average velocity. A common exam trap: "A car travels 100 km in 2 hours. What is its velocity at the 1-hour mark?" The answer is not 50 km/h — that's the average. You need more information (or a position function) to find the instantaneous value.

Important

Instantaneous velocity is a vector — it has both magnitude (speed) and direction. If the object reverses direction, the instantaneous velocity changes sign. The speedometer only shows magnitude.

Why This Matters

Instantaneous velocity is the foundation of all of kinematics and dynamics. Newton's second law ( $F = ma$ ) uses acceleration, which is the instantaneous rate of change of velocity. Without this concept, you can't describe motion that changes — which is almost all real motion.

When you see a graph of position vs. time, the instantaneous velocity at any point is the slope of the tangent line at that point. That geometric interpretation will serve you well in both physics and calculus.

Instantaneous Velocity bridges the NCERT Class 11 Physics chapter on Motion in a Straight Line with the calculus taught in Class 11 Mathematics' Limits and Derivatives chapter, matching searches like "instantaneous velocity: definition and formula" or "kinematics important questions class 11 physics". This distinction from average velocity is a classic conceptual question in CBSE boards and a common numerical setup for JEE Main and NEET.

Yes, this is possible, because velocity and acceleration are two independent instantaneous quantities — one being momentarily zero does not force the other to be zero as well.

Example: a ball thrown vertically upward. Consider a ball thrown straight up with some initial speed. As it rises, gravity continuously decelerates it (acceleration $= g$ , directed downward, throughout the entire flight — this never switches off). At the very instant the ball reaches its highest point, its velocity is exactly zero (it is momentarily neither rising nor falling), yet the acceleration at that same instant is still $g$ , directed downward, exactly as it was a moment before and a moment after. If the acceleration were actually zero at that instant, the ball would remain suspended there forever with zero velocity; instead, it immediately begins to fall, which is only possible because a nonzero downward acceleration was acting on it even while its velocity was zero.

General principle. Velocity, $v = dx/dt$ , and acceleration, $a = dv/dt$ , are both instantaneous rates evaluated at a single moment; there is no mathematical requirement that one vanishing forces the other to vanish. A body can be momentarily at rest (an instant where the position-time graph's tangent is horizontal) while its velocity-time graph is still sloped (nonzero acceleration) at that exact same instant.

[!ANSWER] Yes — a ball thrown vertically upward has zero velocity at its highest point but a nonzero downward acceleration (g) at that same instant, since velocity and acceleration are independent instantaneous quantities.

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

CBSE 2026Set ANNUAL1 markMCQ
$Q. The position of an object moving along x-axis is given by the equation x = 5t - t^2 where x is in m and t in s. The instantaneous speed at t = 2s is :$
(a) 6 m/s
(b) 1 m/s
(c) 14 m/s
(d) 3 m/s
›Reveal solutionSolution

Differentiating x = 5t - t^2 gives v = 5 - 2t; substituting t = 2 s gives instantaneous speed = 1 m/s.

Given: x = 5t - t^2 (x in m, t in s).

Instantaneous velocity is defined as v = dx/dt.

Differentiating:

v = d/dt (5t - t^2) = 5 - 2t

At t = 2 s:

v = 5 - 2(2) = 5 - 4 = 1 m/s

Since this is positive, the instantaneous speed equals the magnitude of velocity, i.e. 1 m/s.

✓Final answer
$The correct option is (b) 1 m/s.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. If the displacement-time graph of a particle is represented by y = mx + c, then the particle is moving with (A) Constant speed (B) Constant velocity (C) Variable velocity (D) Constant momentum$
›Reveal solutionSolution

A linear $x$ - $t$ graph $y=mx+c$ means constant velocity.

Velocity is the slope of the displacement-time graph, $v=dy/dx$ (treating $x$ as time here). For $y=mx+c$ , the slope is $m$ everywhere — it never changes with time, so the particle moves with constant velocity (not merely constant speed, since direction is also fixed along the line).

✓Final answer
$(B) Constant velocity.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. The motion of a particle is described by the equation x = at + bt^2, where a = 15 cms^-1 and b = 3 cms^-2. Its instantaneous velocity at t = 3 s will be$
(1) 33 cms^-1
(2) 18 cms^-1
(3) 16 cms^-1
(4) 32 cms^-1
›Reveal solutionSolution

Instantaneous velocity is v = dx/dt; differentiating the given position equation and substituting t = 3 s gives 33 cm/s.

Given: x = at + bt^2, with a = 15 cm s^-1, b = 3 cm s^-2.

Instantaneous velocity:

v = dx/dt = a + 2bt

At t = 3 s:

v = 15 + 2(3)(3) = 15 + 18 = 33 cm s^-1

✓Final answer
$(1) 33 cm s^-1.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Slope of displacement-time graph in straight line motion represents:$
(a) time
(b) distance
(c) velocity
(d) acceleration
›Reveal solutionSolution

The slope of an x-t graph is $dx/dt$ , which is exactly how velocity is defined.

For motion in a straight line, displacement $x$ is plotted against time $t$ . Velocity is defined as the rate of change of displacement:

$v = \frac{dx}{dt}$

Graphically, $dx/dt$ at any point of the curve is the slope of the tangent to the $x$ - $t$ graph at that point. For uniform motion the graph is a straight line and its constant slope gives the (constant) velocity; for non-uniform motion the slope changes from point to point and gives the instantaneous velocity at each instant. Acceleration, by contrast, is the slope of a velocity-time graph, not a displacement-time graph.

✓Final answer
$The slope of a displacement-time graph represents velocity — option (c).$
CBSE 2021Set sz1 mark
$Q. Define instantaneous velocity.$
›Reveal solutionSolution

Instantaneous velocity is the limiting value of average velocity as the time interval considered tends to zero; it equals dx/dt.

Average velocity over a time interval delta-t is defined as delta-x / delta-t, the displacement divided by the time taken. As delta-t is made smaller and smaller (delta-t -> 0), the average velocity approaches a fixed limiting value, which is called the instantaneous velocity at that instant:

v = lim(delta-t -> 0) (delta-x/delta-t) = dx/dt

This is simply the derivative of position (displacement) with respect to time, evaluated at that instant, and it also equals the slope of the tangent to the position-time (x-t) graph at that point.

✓Final answer
$Instantaneous velocity v = lim(delta-t -> 0) (delta-x/delta-t) = dx/dt — the rate of change of position with time at a particular instant.$
CBSE 2020Set ANNUAL1 markMCQ
$Q. The position of an object moving along x-axis is given by, x = a + bt^2 where a = 2m and b = 3 m/s^2 and t is measured in seconds. The velocity at t = 2s is$
(a) 2 m/s
(b) 6 m/s
(c) 12 m/s
(d) 14 m/s
›Reveal solutionSolution

Differentiating x = a + bt^2 gives v = 2bt = 6t (since b = 3), which evaluates to 12 m/s at t = 2 s.

Given: x = a + b t^2, with a = 2 m, b = 3 m/s^2

Velocity: v = dx/dt = 2bt = 2(3)t = 6t

At t = 2 s:

v = 6 x 2 = 12 m/s

✓Final answer
$(c) 12 m/s.$
CBSE 2017Set ANNUAL1 mark
$Q. What does the slope of displacement-time graph represent?$
›Reveal solutionSolution

The slope of an x-t graph at any instant equals the particle's instantaneous velocity at that instant.

For a particle moving along a straight line, if x is its position at time t, the average velocity between two close instants t and t+Δt is v_avg = Δx/Δt, which is exactly the slope of the chord joining the two corresponding points on the x-t graph. As Δt is made smaller and smaller (Δt → 0), the chord becomes the tangent to the curve at that point, and the average velocity becomes the instantaneous velocity: v = dx/dt. So at every point on an x-t graph, the slope of the tangent drawn at that point gives the instantaneous velocity of the particle at that instant — a steep slope means high speed, a horizontal slope (zero slope) means the particle is momentarily at rest, and a negative slope means the particle is moving in the negative direction.

✓Final answer
$The slope of a displacement-time graph gives the (instantaneous) velocity of the particle.$

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Q.Is it possible for a body to have zero velocity at some instant and yet be accelerating at that very instant? Justify your answer with a suitable example.

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