The position of a particle moving along the x-axis is given by $x(t) = 5t^2 - 3t + 2$ (SI units). Find expressions for its velocity and acceleration as functions of time, and calculate their values at $t = 3$ s.

The position of a particle moving along the x-axis – Class 11 Physics

Numerical · Q14

x(t) = 5t^2 - 3t + 2

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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.

Tip

| Quantity | Average rate | Instantaneous rate |

|----------|--------------|-------------------|

| Speed | Total distance / total time | Speedometer reading |

| Growth | Population change over a decade | Population change per year right now |

| Slope | Slope of secant line | Slope of tangent line |

Why this matters for exams

You'll use this idea to:

Find velocity from a position function (derivative of $s(t)$ )
Find acceleration from velocity (derivative of $v(t)$ )
Determine where a function is increasing or decreasing
Solve optimization problems (where rate of change = 0)

The key is always the same: instantaneous rate of change = derivative = limit of difference quotient. Master that one idea, and half of calculus opens up.

Instantaneous Rate of Change is the core idea behind the derivative in the NCERT Class 11 Mathematics chapter on Limits and Derivatives, matching searches like "instantaneous rate of change formula" or "derivatives important questions class 11 class 12 maths". This same concept, applied to position and velocity, is exactly what underlies kinematics questions in Class 11 Physics and JEE Main and NEET numerical problems.

Velocity function. Differentiating $x(t) = 5t^2 - 3t + 2$ once with respect to time (Section 2.4):

v(t) = \frac{dx}{dt} = 10t - 3

Acceleration function. Differentiating $v(t)$ once more (Section 2.5):

a(t) = \frac{dv}{dt} = 10\ \text{m/s}^2

Since this does not depend on $t$ at all, the acceleration is constant — the motion is uniformly accelerated throughout, with $a = 10\ \text{m/s}^2$ at every instant, including $t=3$ s.

Values at $t = 3$ s.

v(3) = 10(3) - 3 = 30 - 3 = 27\ \text{m/s}

a(3) = 10\ \text{m/s}^2 \quad(\text{same as at any other instant, since } a \text{ is constant})

[!ANSWER] At t = 3 s, the velocity is 27 m/s and the acceleration is 10 m/s² (constant for all t).

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: $v=\dfrac{dS}{dt}=a+2bt$ .

Acceleration is the derivative of velocity: $A=\dfrac{dv}{dt}=2b$ , independent of time.

✓Final answer
$2b$
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)

Step 4: Substitute u = x^2 + 5 back in to express the answer purely in terms of x.

dy/dx = 2x[3(x^2+5)^2 + 2]

✓Final answer
$dy/dx = 2x[3(x^2+5)^2 + 2].$
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):

dy/dx = 2u x (-sin x) = 2 cos x x (-sin x) = -2 sin x cos x.

Step 5: Using the identity 2 sin x cos x = sin 2x, this simplifies to dy/dx = -sin 2x.

✓Final answer
$dy/dx = -2 sin x cos x = -sin 2x.$
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:

dy/dx = (1/2) x^((1/2) - 1) = (1/2) x^(-1/2)

dy/dx = 1/(2 x^(1/2)) = 1/(2 sqrt(x))

✓Final answer
$dy/dx = 1/(2 sqrt(x)).$
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

d/dx(9) = 0 (derivative of a constant is zero)

Adding these:

dy/dx = 12x^2 + 14x + 6

✓Final answer
$dy/dx = 12x^2 + 14x + 6.$
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.

dy/dx = 6x + 4

Step 3: Substitute x = 1.

dy/dx = 6(1) + 4 = 10

✓Final answer
$dy/dx = 6x + 4; at x = 1, dy/dx = 10.$
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:

ds/dt = 12t^2 + 0 = 12t^2

(Physically, since s represents position/displacement, ds/dt represents the instantaneous velocity of the particle as a function of time.)

✓Final answer
$ds/dt = 12t^2.$
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

(Physically, since s is position and ds/dt is velocity, the second derivative d^2s/dt^2 represents the instantaneous acceleration of the particle as a function of time -- here it increases linearly with t.)

✓Final answer
$d^2s/dt^2 = 24t.$

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Q.The position of a particle moving along the x-axis is given by x(t)=5t2−3t+2x(t) = 5t^2 - 3t + 2x(t)=5t2−3t+2 (SI units). Find expressions for its velocity and acceleration as functions of time, and calculate their values at t=3t = 3t=3 s.

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Q.The position of a particle moving along the x-axis is given by $x(t) = 5t^2 - 3t + 2$ (SI units). Find expressions for its velocity and acceleration as functions of time, and calculate their values at $t = 3$ s.

Unlock everything free for 14 days