The displacement (in meter) of a particle moving along x-axis is given by x = t² + 6t. Calculate instantaneous

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Q. The displacement (in meter) of a particle moving along x-axis is given by x = t² + 6t. Calculate instantaneous velocity at t=2s.

Karnataka PUCKarnataka School Examination and Assessment Board (1st PUC) 2025Subjective· 2mImportance★★★★★est

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Concept understanding — Instantaneous Rate Of Change

Instantaneous Rate of Change

Imagine you're in a car watching the speedometer. It doesn't say "I travelled 60 km in the last hour" — it shows your speed right now, at this exact moment. That number, the one that changes every time you tap the brake or press the accelerator, is the instantaneous rate of change of your position.

The intuition: from average to instant

If you drive from Delhi to Agra (200 km) in 4 hours, your average speed is 50 km/h. But that tells you nothing about how fast you were going at 10:15 AM when you passed a particular toll booth. You might have been doing 80 km/h, or 20 km/h if there was traffic.

The average rate of change over a time interval $[t_1, t_2]$ is:

\text{Average speed} = \frac{\text{distance travelled}}{\text{time taken}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}

where $s(t)$ is your position at time $t$ .

To get the speed at a specific moment $t = a$ , you'd want to look at smaller and smaller intervals around $a$ . If you measure from $t = a$ to $t = a + h$ , where $h$ is a tiny time difference:

\text{Average speed over } [a, a+h] = \frac{s(a+h) - s(a)}{h}

As $h$ gets smaller — say 0.1 seconds, then 0.01, then 0.0001 — this average speed gets closer and closer to a single number. That limiting number is the instantaneous rate of change at $t = a$ .

Note

This is the core idea: instantaneous rate of change = the limit of average rates of change as the interval shrinks to zero.

The precise definition

For any function $y = f(x)$ , the instantaneous rate of change at $x = a$ is:

\text{Instantaneous rate of change} = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

provided this limit exists. This limit is also called the derivative of $f$ at $a$ , denoted $f'(a)$ or $\frac{dy}{dx}\big|_{x=a}$ .

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

What it means geometrically

If you graph $y = f(x)$ , the average rate of change over $[a, a+h]$ is the slope of the secant line through $(a, f(a))$ and $(a+h, f(a+h))$ . As $h \to 0$ , that secant line pivots and approaches a tangent line at $x = a$ . The slope of that tangent line is exactly $f'(a)$ .

So instantaneous rate of change = slope of the tangent line.

A concrete example

Let $f(x) = x^2$ . Find the instantaneous rate of change at $x = 3$ .

First, the average rate over $[3, 3+h]$ :

\frac{f(3+h) - f(3)}{h} = \frac{(3+h)^2 - 9}{h} = \frac{9 + 6h + h^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h

Now take the limit as $h \to 0$ :

\lim_{h \to 0} (6 + h) = 6

So at $x = 3$ , the function $x^2$ is changing at a rate of 6 units per unit change in $x$ . The tangent line at $(3, 9)$ has slope 6. …

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[!TLDR]

v = 10 m/s

Method

v = dx/dt = 2t + 6. At t=2s: v = 2(2)+6 …

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

CBSE 2024Set ANNUAL1 markMCQ
$Q. The equation for displacement of a body is given by S = at + bt^2. The acceleration of the body is (A) a/b (B) 2b (C) a + b (D) 3b$
›Reveal solutionSolution

For $S=at+bt^2$ , acceleration is constant and equal to $2b$ .

Velocity is the first derivative: $v=\dfrac{dS}{dt}=a+2bt$ .

…

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CBSE 2023Set annual1 mark
$Q. If y = u^3 + 2u and u = x^2 + 5, find dy/dx.$
›Reveal solutionSolution

Apply the chain rule: dy/dx = (dy/du) x (du/dx).

Given y = u^3 + 2u and u = x^2 + 5.

Step 1: Differentiate y with respect to u.

dy/du = 3u^2 + 2

Step 2: Differentiate u with respect to x.

du/dx = 2x

Step 3: Apply the chain rule.

dy/dx = (dy/du) x (du/dx) = (3u^2 + 2)(2x)

…

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CBSE 2021Set sz1 mark
$Q. If y = cos^2 x, then dy/dx is .............$
›Reveal solutionSolution

Using the chain rule on y = (cos x)^2 gives dy/dx = -2 sin x cos x, which can be written as -sin 2x.

Step 1: Write y = (cos x)^2 and let u = cos x, so y = u^2.

Step 2: Differentiate y with respect to u: dy/du = 2u.

Step 3: Differentiate u with respect to x: du/dx = -sin x.

Step 4: Apply the chain rule, dy/dx = (dy/du)(du/dx): …

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CBSE 2020Set annual1 mark
$Q. If y = sqrt(x), find dy/dx.$
›Reveal solutionSolution

For y = sqrt(x) = x^(1/2), the derivative is dy/dx = 1/(2 sqrt(x)), obtained by the power rule.

Given y = sqrt(x), rewrite it as y = x^(1/2).

Using the standard power-rule result d(x^n)/dx = n x^(n-1) with n = 1/2:

…

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CBSE 2020Set hz1 mark
$Q. Find dy/dx, when y = 4x^3 + 7x^2 + 6x + 9$
›Reveal solutionSolution

Differentiate each term of the polynomial using the power rule d/dx(x^n) = n x^(n-1).

Given y = 4x^3 + 7x^2 + 6x + 9.

Differentiate term by term:

d/dx(4x^3) = 4 * 3x^2 = 12x^2

d/dx(7x^2) = 7 * 2x = 14x

d/dx(6x) = 6 …

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CBSE 2019Set annual1 mark
$Q. If y = 3x^2 + 4x + 5, find dy/dx, at x = 1.$
›Reveal solutionSolution

Differentiate each term of y = 3x^2 + 4x + 5 using the power rule; dy/dx = 6x + 4, so at x = 1 the derivative is 10.

Step 1: Differentiate term by term.

d/dx(3x^2) = 3 * 2x = 6x

d/dx(4x) = 4

d/dx(5) = 0 (derivative of a constant is zero)

Step 2: Add the results. …

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CBSE 2018Set annual1 mark
$Q. Given s = 4t^3 + 3, calculate ds/dt.$
›Reveal solutionSolution

Differentiating s = 4t^3 + 3 with respect to t (using the power rule) gives ds/dt = 12t^2.

Given: s = 4t^3 + 3

Differentiate each term with respect to t. Using the power rule, d(t^n)/dt = n t^(n-1):

d(4t^3)/dt = 4 x 3 x t^(3-1) = 12 t^2

The derivative of the constant term (3) with respect to t is zero, since a constant does not change with time:

d(3)/dt = 0

Adding the two results:

…

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CBSE 2018Set annual1 mark
$Q. Given s = 4t^3 + 3, calculate d^2s/dt^2.$
›Reveal solutionSolution

Differentiating ds/dt = 12t^2 once more with respect to t gives the second derivative, d^2s/dt^2 = 24t.

From the previous part, ds/dt = 12t^2. The second derivative d^2s/dt^2 is obtained by differentiating this expression again with respect to t, using the power rule:

d^2s/dt^2 = d(12t^2)/dt = 12 x 2 x t^(2-1) = 24t

…

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Q.The displacement (in meter) of a particle moving along x-axis is given by x = t² + 6t. Calculate instantaneous velocity at t=2s.

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