Formula of Motion: The Three Equations of Motion Explained

The formula of motion that students learn first is really a set of three equations: first $v = u + at$ , second $s = ut + \frac{1}{2}at^2$ , third $v^2 = u^2 + 2as$ . Together they describe any object moving in a straight line with uniform acceleration, from a car pulling away at a traffic light to a stone dropped down a well.

There are five quantities in play (starting velocity, final velocity, acceleration, displacement and time), and each equation uses four of them while ignoring one. So the skill is not memorising the equations, which takes a minute, but spotting which quantity a question does not mention and picking the equation that does without it. These equations come up in Class 9 motion, Class 11 straight line motion and the opening kinematics unit of AP Physics 1, where they are called the kinematic equations.

Quick facts

First equation
$v = u + at$
Second equation
$s = ut + \frac{1}{2}at^2$
Third equation
$v^2 = u^2 + 2as$
Average velocity form
$s = \frac{(u + v)}{2}t$
Condition
Straight line, constant acceleration
Free fall
Replace a with g = 9.8 m/s²

Equation of motion formula: which one to use

Each equation of motion and the quantity it leaves out
Equation	Links	Leaves out
$v = u + at$	u, v, a, t	s
$s = ut + \frac{1}{2}at^2$	s, u, a, t	v
$v^2 = u^2 + 2as$	u, v, a, s	t
$s = \frac{(u + v)}{2}t$	s, u, v, t	a

Write down the three values you are given and the one you want. The fifth quantity, the one that is neither given nor asked, tells you which row to use. That single habit solves most motion questions in one line.

Deriving the equations from a velocity time graph

Class 9 asks for the graphical derivation. Draw a straight line on a velocity time graph from u at t = 0 to v at time t.

First equation. The slope of the line is the acceleration, so $a = \frac{v - u}{t}$ , which rearranges to $v = u + at$ .
Second equation. The region enclosed below the line equals the displacement. That region is a rectangle of area ut plus a triangle of area $\frac{1}{2} \times t \times at$ , giving $s = ut + \frac{1}{2}at^2$ .
Third equation. The same area is a trapezium, $s = \frac{(u + v)}{2}t$ . Substitute $t = \frac{v - u}{a}$ and simplify: $2as = v^2 - u^2$ .

Each step uses only the slope and the area, so if you can read those two features of a graph, you can rebuild the equations in an exam even if you forget them.

Physics motion formulas in action

Example 1: a train leaving a station

A metro train pulls out from rest, gaining speed steadily at 0.50 m/s² for 1.0 min. Find its final velocity and the distance covered.

t = 60 s. $v = 0 + 0.50 \times 60 = 30\ \text{m/s}$ .

$s = 0 + \frac{1}{2}(0.50)(60)^2 = 900\ \text{m}$ .

Check with the third equation: $v^2 = 2(0.50)(900) = 900$ , so v = 30 m/s.

Example 2: a bus braking, speed in km/h

A bus moving at 54 km/h brakes uniformly and stops in 45 m. Find the retardation and the stopping time.

u = 54 ÷ 3.6 = 15 m/s, v = 0, s = 45 m. Time is not given, so use the third equation.

$0 = 15^2 + 2a(45) \Rightarrow a = -\frac{225}{90} = -2.5\ \text{m/s}^2$ . The retardation is 2.5 m/s².

$t = \frac{v - u}{a} = \frac{0 - 15}{-2.5} = 6.0\ \text{s}$ .

Example 3: a stone off a bridge

A stone dropped from a bridge hits the water 2.5 s later. How high is the bridge, and how fast does the stone hit? Ignore air resistance.

u = 0, a = g = 9.8 m/s². $s = \frac{1}{2}(9.8)(2.5)^2 = 4.9 \times 6.25 \approx 30.6\ \text{m}$ .

$v = 0 + 9.8 \times 2.5 = 24.5\ \text{m/s}$ .

Formula of motion class 9: the distance in the nth second

Class 9 and Class 11 papers sometimes ask for the distance covered during one particular second, the nth second. Subtract the displacement after n − 1 seconds from the displacement after n seconds:

$s_{n\text{th}} = u + a\left(n - \tfrac{1}{2}\right)$

For a body falling from rest, the distances in the first, second and third seconds are 4.9 m, 14.7 m and 24.5 m, in the ratio 1 : 3 : 5. This odd number pattern is a quick check on free fall answers.

When the equations do not apply

Changing acceleration. A rocket burning fuel or a car with varying throttle needs calculus or a graph.
Curved paths. Use the equations separately along each axis, as in projectile motion.
Air resistance that matters. A falling feather does not have constant acceleration.

Related pages: final velocity formula, formula of displacement, constant acceleration formula and acceleration formula. The kinematics formulas hub also has the projectile versions.

Exam slips

Mixed signs. Pick a positive direction and give u, v, a and s signs to match it.
Unconverted speeds. km/h must become m/s before you use m/s² values.
Half forgotten. The second equation has ½at², and the third has 2as.
Square root dropped. The third equation gives v², not v.

Frequently asked questions

What are the three equations of motion?

First, v = u + at. Second, s = ut + ½at². Third, v² = u² + 2as. All three hold for straight line motion with constant acceleration.

What is the formula of motion for Class 9?

Class 9 uses the same three equations of motion, derived from a velocity time graph. A common extra is the distance covered in the nth second, u + a(n − ½).

Which equation of motion has no time in it?

The third one, v² = u² + 2as. Use it when the question gives distance but not time.

Can the equations of motion be used for free fall?

Yes. Replace a with g = 9.8 m/s² (downward), as long as air resistance is small.

Why do the equations of motion need constant acceleration?

They are derived from a straight line velocity time graph. If acceleration changes, the graph curves and the slope and area rules no longer give these simple forms.