A particle moves along the x-axis in such a way that its coordinates x varies with time 't' according to equat

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Q. A particle moves along the x-axis in such a way that its coordinates x varies with time 't' according to equation x = 2 - 5t + 6t^2. What is the initial velocity of the particle ?

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2022Subjective· 2mImportance★★★★★

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

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Differentiate the given position-time equation once with respect to time to get the velocity-time equation, then plug in t = 0 to read off the initial velocity.

Given position equation:

x = 2 − 5t + 6t² (in appropriate SI units)

Step 1 — velocity is the time derivative of position:

v = dx/dt = d/dt [2 − 5t + 6t²]

v = 0 − 5 + 12t

v = -5 + 12t

Step 2 — find the INITIAL velocity by substituting t = 0:

v(t=0) = -5 + 12(0) = -5 ms^-1

…

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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

…

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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 velo …

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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: …

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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 give …

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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

…

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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

…

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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 …

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Q.A particle moves along the x-axis in such a way that its coordinates x varies with time 't' according to equation x = 2 - 5t + 6t^2. What is the initial velocity of the particle ?

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