JEE Mains Physics · Motion in a Straight Line

Equations of Uniformly Accelerated Motion

Under constant acceleration, v = u + at, s = ut + ½at² and v² = u² + 2as; pick the one that leaves out the quantity you are neither given nor asked for.

All Motion in a Straight Line notes JEE Mains Physics strategy

Why this matters

Twenty-one PYQs, nine of them asking for a number, and two from 2026. Nine apply the equations to one stretch of motion, six are about braking and stopping distances, and six about the distance covered in one particular second. The algebra is short; what costs marks is a sign on a retardation or an "in the nth second" read as "in n seconds".

Concept 1 of 3: Choosing the right equation of motion

Each equation links four of the five quantities u, v, a, s and t. List what is given and what is asked; the one quantity that is neither tells you which equation to use. When a body speeds up and then slows down, draw the velocity–time graph: it is a triangle, and its area is the distance.

Definition

No s: $v = u + at$ . No v: $s = ut + \tfrac{1}{2}at^{2}$ . No t: $v^{2} = u^{2} + 2as$ . No a: $s = \dfrac{u + v}{2}\,t$ .
Speed at the MIDDLE of a distance (the middle of a train passing a post): $v_m = \sqrt{\dfrac{u^{2} + v^{2}}{2}}$ . Speed at the middle of the TIME: $\dfrac{u + v}{2}$ .
Speeding up at $\alpha$ from rest, then braking at $\beta$ to rest, total time t: $v_{max} = \dfrac{\alpha\beta t}{\alpha + \beta}$ , distance $= \tfrac{1}{2}v_{max}t = \dfrac{\alpha\beta t^{2}}{2(\alpha + \beta)}$ .
In that case $\dfrac{t_1}{t_2} = \dfrac{\beta}{\alpha}$ : the same speed is gained and lost.
A constant force from rest: $a = F/m$ along each axis, then $s = \tfrac{1}{2}at^{2}$ along each axis.
Sliding up a smooth incline: $a = -g\sin\theta$ , so it stops after $\dfrac{u^{2}}{2g\sin\theta}$ .

Equations of motion

v = u + at \qquad s = ut + \tfrac{1}{2}at^{2} \qquad v^{2} = u^{2} + 2as \qquad s = \frac{u + v}{2}\,t

Worked example

A car speeds up uniformly from

10\ \text{m/s}

30\ \text{m/s}

over 200 m. Find the acceleration, the time taken and the speed at the halfway mark.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 3 · Q8Moderate

Example 1 · Motion in a Straight Line · Equations of Uniformly Accelerated Motion

An engine of a train, moving with uniform acceleration, passes the signal post with velocity

u

and the last compartment with velocity

v

. The velocity with which middle point of the train passes the signal post is:

Mid-distance speed is not the mean speed

The middle of a train passes the post when half the LENGTH has gone by, so its speed is √((u² + v²)/2). The mean (u + v)/2 is the speed at half the TIME.

The ratio of times is upside down

Speeding up at a₁ and braking at a₂ over the same speed change gives t₁/t₂ = a₂/a₁. The larger acceleration takes the shorter time.

Drill 8 more on choosing the right equation of motion

Concept 2 of 3: Braking and stopping distance

With constant braking, the stopping distance is u²/2a. It grows with the SQUARE of the speed: twice the speed needs four times the distance. When a body loses part of its speed over a known distance, write v² = u² − 2as for that part and for the rest, and divide; the retardation cancels.

Definition

Stopping distance $s = \dfrac{u^{2}}{2a}$ ; stopping time $t = \dfrac{u}{a}$ .
Same retardation: $s \propto u^{2}$ . One third of the speed stops in one ninth of the distance.
Distance in a uniform stop of known time: $s = \dfrac{u}{2}\,t$ .
A bullet slowing from u to ku over $d_1$ stops after a further $d_2$ : $\dfrac{d_2}{d_1} = \dfrac{k^{2}}{1 - k^{2}}$ .
Braking applied late: $v^{2} = u^{2} - 2as$ with the shorter s gives the speed left over.
Two cars braking towards each other: each stops after its own $u^{2}/2a$ ; subtract both from the gap.

Stopping distance

s = \frac{u^{2}}{2a} \qquad \frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^{2}

Worked example

A car at

72\ \text{km/h}

stops in 40 m. Find its retardation, and the stopping distance at

108\ \text{km/h}

with the same brakes.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q110Moderate

Example 2 · Motion in a Straight Line · Equations of Uniformly Accelerated Motion

A bullet is fired into a fixed target loses one third of its velocity after travelling 4 cm. It penetrates further

D \times10^{- 3}\text{ }m

before coming to rest. The value of

D

is:

Stopping distance is not proportional to speed

Halving the speed quarters the stopping distance. Scaling the distance by the speed ratio alone is the usual wrong option.

Loses one third, or keeps one third?

"Loses one third of its velocity" leaves 2u/3. "Velocity becomes one third" leaves u/3. The two give very different answers.

Drill 5 more on braking and stopping distance

Concept 3 of 3: Distance in the nth second

The distance in the nth second is the distance in n seconds minus the distance in n − 1 seconds. Under constant acceleration it rises by a fixed amount, a, every second. From rest the distances in successive seconds go 1 : 3 : 5 : 7, and the total distance grows as t².

Definition

$s_n = u + \dfrac{a}{2}(2n - 1)$ .
Consecutive seconds differ by a: $s_{n+1} - s_n = a$ ; seconds k apart differ by ka.
From rest: $s_1 : s_2 : s_3 = 1 : 3 : 5$ , and $s \propto t^{2}$ , so the next equal interval covers three times the first.
From rest, $\dfrac{s_{n-1}}{s_n} = \dfrac{2n - 3}{2n - 1}$ .
Displacement $\Delta s$ and velocity gain $\Delta v$ over one second fix the speed at its start: $\Delta s = v + \tfrac{1}{2}\Delta v$ .

Distance in the nth second

s_n = u + \frac{a}{2}(2n - 1)

Worked example

A body covers 20 m in its 3rd second and 32 m in its 5th second, with constant acceleration. Find u and a.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q14Moderate

Example 3 · Motion in a Straight Line · Equations of Uniformly Accelerated Motion

A body travels

102.5\text{ }m

n^{\text{th~}}

second and

115.0\text{ }m

(n + 2)^{\text{th~}}

second. The acceleration is :

"In the nth second" is not "in n seconds"

In the 5th second means between t = 4 s and t = 5 s. In 5 seconds means from t = 0 to t = 5 s. Read the stem twice.

Seconds two apart differ by 2a

The nth and (n + 2)th seconds differ by 2a, not a. Dividing the difference by 1 doubles the acceleration.

Drill 5 more on distance in the nth second

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)

Choosing the right equation of motion
Equations of motion
$v = u + at \qquad s = ut + \tfrac{1}{2}at^{2} \qquad v^{2} = u^{2} + 2as \qquad s = \frac{u + v}{2}\,t$
Braking and stopping distance
Stopping distance
$s = \frac{u^{2}}{2a} \qquad \frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^{2}$
Distance in the nth second
Distance in the nth second
$s_n = u + \frac{a}{2}(2n - 1)$

Watch out for (6)

Mid-distance speed is not the mean speed
→ Choosing the right equation of motion
The ratio of times is upside down
→ Choosing the right equation of motion
Stopping distance is not proportional to speed
→ Braking and stopping distance
Loses one third, or keeps one third?
→ Braking and stopping distance
"In the nth second" is not "in n seconds"
→ Distance in the nth second
Seconds two apart differ by 2a
→ Distance in the nth second

Test yourself on Motion in a Straight Line

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