JEE Mains Physics · Motion in a Straight Line

Variable Acceleration: Differentiate and Integrate

When the acceleration is not constant, the equations of motion fail: differentiate a position to get velocity and acceleration, use a = v dv/dx when velocity is given in terms of position, and integrate an acceleration with the starting values to get back velocity and position.

All Motion in a Straight Line notes JEE Mains Physics strategy

Why this matters

Twenty-three PYQs, six of them asking for a number, and two from 2026. Twelve give the position as a function of time and ask for a velocity, an acceleration or a turning point; seven give the velocity in terms of position; four integrate an acceleration, a velocity or a force. Recognising which of the three is given is most of the work.

Concept 1 of 3: Position as a function of time: differentiate

If x(t) is given, velocity is its derivative and acceleration is the derivative of that. A turning point is where the velocity is zero, because that is where the body reverses. "Velocity when the acceleration is zero" means: solve a = 0 for t, then put that t into v. Vectors are handled one component at a time.

Definition

$v = \dfrac{dx}{dt}$ , $a = \dfrac{dv}{dt} = \dfrac{d^{2}x}{dt^{2}}$ .
Turning point: $v = 0$ . Velocity when $a = 0$ : solve $a = 0$ , substitute in v.
Distance over an interval with a turning point: add the sizes of the displacements on each side.
Implicit relations such as $x^{2} = c + t^{2}$ : differentiate both sides, $xv = t$ , then again, $v^{2} + xa = 1$ .
$\vec r = x(t)\hat i + y(t)\hat j$ : $\vec v$ and $\vec a$ component by component; force $= m\vec a$ .
Given t as a function of x: $\dfrac{dx}{dt} = 1 \big/ \dfrac{dt}{dx}$ .

Differentiate

v = \frac{dx}{dt} \qquad a = \frac{dv}{dt} = \frac{d^{2}x}{dt^{2}}

Worked example

x = t^{3} - 9t^{2} + 24t

(x in m, t in s). Find when the particle turns, its velocity when the acceleration is zero, and the distance covered in the first 4 s.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q8Moderate

Example 1 · Motion in a Straight Line · Variable Acceleration: Differentiate and Integrate

A particle starts moving from time

t = 0

and its coordinate is given as

x(t) = 4t^{3}- 3t

. (A) The particle returns to its original position (origin) 0.866 units later (B) The particle is 1 unit away from origin at its turning point. (C) Acceleration of the particle is non-negative. (D) The particle is 0.5 units away from origin at its turning point. (E) Particle never turns back as acceleration is non-negative. Choose the correct answer from the options given below:

A turning point is v = 0, not a = 0

The body reverses where its velocity changes sign. Where a = 0 the velocity is largest or smallest, which is a different instant.

Distance across a turning point

If the body turns inside the interval, x(end) − x(start) is only the displacement. Split at the turning point and add the sizes.

Drill 11 more on position as a function of time: differentiate

Concept 2 of 3: Velocity as a function of position: a = v dv/dx

When v is given in terms of x, the chain rule turns dv/dt into v dv/dx. A velocity that grows as the square root of x gives a constant acceleration, which is why v = k√x keeps appearing: it is uniform acceleration from rest in disguise. When time is given in terms of position, flip it first: v = 1/(dt/dx).

Definition

$a = \dfrac{dv}{dt} = v\dfrac{dv}{dx}$ .
$v = k\sqrt{x}$ : $v^{2} = k^{2}x$ , so $a = \dfrac{k^{2}}{2}$ , constant; force $= \dfrac{mk^{2}}{2}$ .
$v = \sqrt{c + bx}$ : $a = \dfrac{b}{2}$ .
"Velocity grows by k per metre" means $\dfrac{dv}{dx} = k$ , so $a = kv$ .
$t = \alpha x^{2} + \beta x$ : $v = \dfrac{1}{2\alpha x + \beta}$ and $a = -2\alpha v^{3}$ .
A velocity field $v_x(x)$ : $a_x = v_x\dfrac{\partial v_x}{\partial x}$ , one component at a time.

Chain rule

a = v\frac{dv}{dx} \qquad v = k\sqrt{x} \Rightarrow a = \frac{k^{2}}{2}

Worked example

A 2 kg body moves along x with

v = 6\sqrt{x}

(v in m/s, x in m). Find its acceleration and the force on it.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q3Moderate

Example 2 · Motion in a Straight Line · Variable Acceleration: Differentiate and Integrate

The relation between time '

t

' and distance '

x

' is

t =\alpha x^{2}+ \beta x

, where

\alpha

and

\beta

are constants. The relation between acceleration (a) and velocity (v) is:

dv/dx is not the acceleration

"Velocity increases at 5 m/s per metre" is dv/dx. The acceleration is v times that, so it depends on where the body is.

Inverting the derivative but not the function

From t(x), v is 1/(dt/dx), not dt/dx. Check the units: dt/dx is in seconds per metre.

Drill 6 more on velocity as a function of position: a = v dv/dx

Concept 3 of 3: Acceleration or force as a function of time: integrate

Going the other way, integrate the acceleration to get the velocity and integrate again to get the position. Each integration needs a starting value: the initial velocity and the initial position. A force that varies with time does the same job through a = F/m, and its area against time is the change in momentum.

Definition

$v = u + \displaystyle\int_0^{t} a\,dt$ , $x = x_0 + \displaystyle\int_0^{t} v\,dt$ .
Displacement between $t_1$ and $t_2$ : $\displaystyle\int_{t_1}^{t_2} v\,dt$ .
$F(t)$ : $m\Delta v = \displaystyle\int F\,dt$ , the area under F–t.
If v changes sign inside the interval, the integral is the displacement; the distance needs the parts added as sizes.
Two bodies, one with $a \propto t$ and one with constant a: their gap is a cubic in t, so they meet at most three times.

Integrate with the starting values

v(t) = u + \int_0^{t} a\,dt \qquad x(t) = x_0 + \int_0^{t} v\,dt

Worked example

A particle at x = 0 moving at

2\ \text{m/s}

has acceleration

a = 6t\ \text{m/s}^{2}

. Find its velocity and position at

t = 2\ \text{s}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · 2021 Compilation Paper 15 · Q3Moderate

Example 3 · Motion in a Straight Line · Variable Acceleration: Differentiate and Integrate

A particle starts from rest with acceleration

a=\alpha t+\beta t^{2}

where

\alpha

and

\beta

are constant. Find its displacement between

t = 1

and

t = 2

seconds.

Dropping the starting value

An integral fixes the change, not the value. Leaving out the initial velocity or position shifts every answer that follows.

Differentiating when you should integrate

If v(t) is given and a displacement is asked, integrate v. Differentiating gives the acceleration, which is often among the options.

Drill 3 more on acceleration or force as a function of time: integrate

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Position as a function of time: differentiate
Differentiate
$v = \frac{dx}{dt} \qquad a = \frac{dv}{dt} = \frac{d^{2}x}{dt^{2}}$
Velocity as a function of position: a = v dv/dx
Chain rule
$a = v\frac{dv}{dx} \qquad v = k\sqrt{x} \Rightarrow a = \frac{k^{2}}{2}$
Acceleration or force as a function of time: integrate
Integrate with the starting values
$v(t) = u + \int_0^{t} a\,dt \qquad x(t) = x_0 + \int_0^{t} v\,dt$

Watch out for (6)

A turning point is v = 0, not a = 0
→ Position as a function of time: differentiate
Distance across a turning point
→ Position as a function of time: differentiate
dv/dx is not the acceleration
→ Velocity as a function of position: a = v dv/dx
Inverting the derivative but not the function
→ Velocity as a function of position: a = v dv/dx
Dropping the starting value
→ Acceleration or force as a function of time: integrate
Differentiating when you should integrate
→ Acceleration or force as a function of time: integrate

Test yourself on Motion in a Straight Line

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