Equations of Uniformly Accelerated Motion

The three kinematic equations v = u + at, s = ut + ½at² and v² = u² + 2as are the skeleton of 1D kinematics. Learn when each applies — and the sign conventions that decide whether you get the answer right.

Equations of Motion1 min · Free lecture

In this lesson

State the three equations of uniformly accelerated motion and their conditions
Apply sign conventions for displacement, velocity and acceleration
Solve free-fall problems including the nth-second formula

For motion in a straight line with constant acceleration $a$ , initial velocity $u$ , and time $t$ , three equations describe everything. They hold only when $a$ is constant — that is their one condition, and JEE loves testing it.

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

The three kinematic equations

u — initial velocity (m/s)
v — velocity after time t (m/s)
a — constant acceleration (m/s^2)
s — displacement in time t (m)

Sign convention is everything. Choose a positive direction once, then u, v, a and s all carry signs. In free fall, a = −g when upward is positive — students who drop the minus sign lose the answer by 2×.

A frequent exam favourite: displacement in the $n$ th second. The distance covered between $t = n-1$ and $t = n$ seconds is the difference of displacements — it comes out to a clean formula.

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

Displacement in the nth second (uniformly accelerated motion)

s_n — displacement during the nth second (m)
n — the nth second of motion

Worked example

A body starts from rest and accelerates uniformly at 2 m/s². Find (a) its velocity after 5 s, (b) the displacement in the 5th second.

(a) v = u + at = 0 + 2 × 5 = 10 m/s
(b) s_5 = u + (a/2)(2n − 1) = 0 + 1 × 9 = 9 m

Answer: v = 10 m/s; displacement in 5th second = 9 m

Animated figure: velocity-time-graph (rendered in video)

Velocity-time graph: slope is acceleration, area under the curve is displacement

Transcript (1 min)

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Equations of Uniformly Accelerated Motion
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v = u + at · s = ut + ½ at² · v² = u² + 2as
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v–t graph: slope = a, area = s
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Sign conventions decide the answer
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s_n = u + (a/2)(2n − 1)
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Example: u = 0, a = 2 m/s² → v(5s) = 10 m/s, s₅ = 9 m
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Recap: constant a → three equations + graphs

Frequently asked

When can I NOT use the three kinematic equations?

They are valid only for constant acceleration. If acceleration varies with time or position (like a spring force), you must integrate a = dv/dt instead — the equations give wrong answers outside constant-a motion.

How do I choose the sign of g in free-fall problems?

Pick a positive direction once. If upward is positive, then for a falling body a = −g; if downward is positive, a = +g. The equations themselves are sign-consistent — the classic error is mixing conventions mid-problem.

Printable notes (free)

Practice

3 free · 1 premium

Q1 · Numerical · difficulty 1/5

A body moving with initial velocity 10 m/s accelerates uniformly at 4 m/s². Find its velocity after 9 s.

Hint available · solution with Premium.

Q2 · Numerical · difficulty 2/5

A body starts from rest and accelerates uniformly at 9 m/s². Find the displacement in the 8th second of its motion.

Hint available · solution with Premium.

Q3 · Numerical · difficulty 2/5

A stone is dropped from rest from a height of 75 m. Taking g = 10 m/s², find the time it takes to reach the ground (neglect air resistance).

Hint available · solution with Premium.

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