Equations of Motion: Three Equations and Their Proofs

A cricket ball leaves the bowler's hand, a train pulls away from a station, a stone drops from a bridge. In each case something moves along a straight line and its speed changes steadily. How fast will it be going after 5 seconds? How far will it have travelled? How long will it take to stop? Three short equations, the equations of motion, answer all these questions. They are among the most useful formulas in school physics, and it is just as important to know where they come from as it is to use them. This article states the equations, proves each one in several different ways (using the definition of acceleration, graphs, algebra, calculus and energy), and then works through many examples, from braking cars to balls thrown upwards.

1.Quantities and Conditions

The equations of motion belong to kinematics, the part of mechanics that describes motion without asking what causes it. They use five quantities, usually remembered as SUVAT:

The three equations are valid only when all of the following hold:

• the object moves in a straight line (one dimension);
• the acceleration is constant (same size and direction throughout);
• s is measured from the starting position, and t from the start of the interval.

The note in brackets is the quantity missing from each equation, which is the key to choosing the right one quickly.

2.Sign Convention

Because u, v, s and a are vectors, signs matter. Choose one direction as positive and keep to it for the whole problem. A common choice is to take the direction of the initial motion as positive. Then a speeding-up object has a positive acceleration, and a slowing-down object has a negative acceleration (deceleration). For vertical motion it is often best to take upward as positive, so that gravity gives a = −g = −9.8 m/s² for every free-falling object, whether it is rising or falling.

3.Equation 1: v = u + at

Proof 1: from the definition of acceleration

Acceleration is the rate of change of velocity. If the acceleration is constant, the velocity changes by the same amount every second, so

a = (change in velocity) ÷ (time taken) = (v − u) ÷ t

Multiply both sides by t: at = v − u. Add u to both sides:

v = u + at   ∎

Proof 2: from the velocity–time graph

For constant acceleration the velocity–time graph is a straight line. The gradient of a velocity–time graph is the acceleration. The line starts at height u (at t = 0) and ends at height v (at time t). Its gradient is the rise divided by the run:

gradient = a = (v − u) ÷ t   ⇒   v = u + at   ∎

In words: the final velocity equals the starting velocity plus the total change, and the total change is acceleration multiplied by time.

4.Equation 2: s = ut + ½at²

Proof 1: area under the velocity–time graph

On a velocity–time graph, the area under the line equals the displacement. (Why? In a short interval of time Δt the object moves v × Δt, which is the area of a thin strip under the graph. Adding up all the strips gives the total displacement.)

The area under the line splits into two simple shapes:

• a rectangle of height u and width t, with area u × t;
• a triangle on top of it, with width t and height (v − u). From Equation 1, v − u = at, so its area is ½ × t × at = ½at².

Adding: s = ut + ½at²   ∎

Proof 2: using average velocity

When velocity changes at a constant rate, the average velocity over the interval is exactly halfway between the starting and finishing values:

v̄ = (u + v) ÷ 2

Displacement equals average velocity multiplied by time, so s = ½(u + v)t. Now replace v by u + at (Equation 1):

s = ½(u + u + at) t = ½(2u + at) t = ut + ½at²   ∎

This "mean speed" idea has an old and interesting history. It was stated by scholars at Merton College, Oxford, in the 14th century (the "Oxford Calculators"), and the French scholar Nicole Oresme gave a graphical proof using areas. Galileo used the same idea in his Two New Sciences (1638) when he analysed uniformly accelerated motion.

5.Equation 3: v² = u² + 2as

Proof 1: eliminating time algebraically

Sometimes we know nothing about the time. We can remove t from the equations. Start with Equation 1 and make t the subject:

t = (v − u) ÷ a

Use the average-velocity form s = ½(u + v)t and substitute:

s = ½(u + v) × (v − u) ÷ a = (v² − u²) ÷ (2a)

because (u + v)(v − u) = v² − u² (the difference of two squares). Multiply by 2a:

2as = v² − u²   ⇒   v² = u² + 2as   ∎

Proof 2: by substitution into Equation 2

Square Equation 1: v² = (u + at)² = u² + 2uat + a²t². Factor the last two terms: v² = u² + 2a(ut + ½at²). The bracket is exactly s (Equation 2), so

v² = u² + 2as   ∎

Proof 3: using work and energy

Suppose a resultant force F acts on a mass m and gives it a constant acceleration a, so F = ma (Newton's second law). Over a displacement s the work done is F × s = mas. By the work–energy principle, this equals the change in kinetic energy:

mas = ½mv² − ½mu²   ⇒   as = ½v² − ½u²   ⇒   v² = u² + 2as   ∎

This proof shows how the kinematics of this article connects with the energy ideas in the articles on work and on energy.

6.The Extra Equations

Two more equations follow from the same ideas and are often useful:

Proof of (5). From Equation 1, u = v − at. Substitute into Equation 2: s = (v − at)t + ½at² = vt − at² + ½at² = vt − ½at²  ∎

7.Calculus Proof (Extension)

If you have met calculus, the equations of motion follow in a few lines. By definition, velocity is the rate of change of displacement, v = ds/dt, and acceleration is the rate of change of velocity, a = dv/dt. With a constant:

• Equation 1: integrate a with respect to time: v = ∫a dt = at + C. At t = 0, v = u, so C = u and v = u + at.
• Equation 2: integrate v with respect to time: s = ∫(u + at) dt = ut + ½at² + C′. At t = 0, s = 0, so C′ = 0 and s = ut + ½at².
• Equation 3: write a = dv/dt = (dv/ds)(ds/dt) = v dv/ds. Then a ds = v dv. Integrate from the start (s = 0, velocity u) to the end (s, velocity v): a s = ½v² − ½u², so v² = u² + 2as.

Notice the same pattern in all the proofs: the algebraic, graphical, energy and calculus methods agree, which is a good sign that the equations are sound.

8.Which Equation Should I Use?

Write down the five SUVAT quantities. Mark the three you know and the one you want. The quantity that is neither known nor wanted is the missing one. Then pick the equation that leaves it out.

Method in five steps: (1) Draw a sketch and choose the positive direction. (2) List s, u, v, a, t with signs and units, converting to SI (for example, km/h to m/s). (3) Identify the unknown and the missing quantity. (4) Choose the equation. (5) Solve, and check that the answer is sensible and has the right unit and sign.

9.What the Equations Tell Us

For motion starting from rest (u = 0) with constant acceleration a, the equations simplify to v = at, s = ½at² and v² = 2as. Three important patterns follow:

• Velocity is proportional to time (v ∝ t): twice as long, twice as fast.
• Distance is proportional to the square of time (s ∝ t²): twice as long, four times as far; three times as long, nine times as far.
• v² is proportional to distance (v² ∝ s): to double the speed you need four times the distance. This is why braking distance grows so quickly with speed.

The odd-number pattern comes straight from Equation 2. The distance moved in the nth second is s(n) − s(n − 1) = ½a[n² − (n − 1)²] = ½a(2n − 1). For a = 2 m/s², the 1st, 2nd, 3rd, 4th and 5th seconds give 1, 3, 5, 7 and 9 m. Galileo discovered this relationship in the early 1600s by timing balls rolling down gentle slopes.

10.Worked Examples

11.Free Fall, Ramps and Projectiles

Free fall. Near the Earth's surface, an object falling under gravity alone has a constant acceleration g ≈ 9.8 m/s² downward. This is why a stone and a hammer fall together in a vacuum. Substituting a = g into the equations gives, for an object dropped from rest, v = gt, s = ½gt² and v² = 2gs. Air resistance is ignored in these formulas; for light, spread-out objects such as feathers or paper, or at high speeds, it cannot be ignored.

Ramps. Galileo studied motion by rolling balls down gentle slopes, which "dilutes" gravity so that the motion is slow enough to time with the instruments of his day. For an ideal frictionless slope at angle θ, the acceleration along the slope is a = g sin θ. For example, on a slope of 30°, a = 9.8 × 0.5 = 4.9 m/s², half of the free-fall value.

Projectiles. The key insight, also due to Galileo, is that horizontal and vertical motions are independent. Horizontally (ignoring air resistance) there is no acceleration, so the velocity is constant. Vertically there is the constant acceleration g, so the equations apply to the vertical motion.

12.When the Equations Do Not Apply

Whenever the acceleration changes, the three equations fail and the motion must be analysed with graphs or calculus instead. Examples include:

• a skydiver approaching terminal velocity, where air resistance increases as speed increases;
• a car whose driver presses and releases the accelerator;
• a spring-driven oscillation, where the force (and so the acceleration) depends on position;
• a rocket, whose mass changes as fuel is burned;
• circular motion, where the direction of the velocity, and so the acceleration, keeps changing.

In such cases, split the motion into intervals where the acceleration is approximately constant, or use the area under a velocity–time graph and the gradient of a displacement–time graph, as described in the article on motion.

13.Common Mistakes to Avoid

14.Check Your Understanding

Try each question before opening the answer.

Summary

• For straight-line motion with constant acceleration: v = u + at, s = ut + ½at², v² = u² + 2as, and the extras s = ½(u + v)t and s = vt − ½at².
• Equation 1 follows from the definition of acceleration (or the gradient of a v–t graph). Equation 2 follows from the area under the v–t graph (a rectangle plus a triangle) or from the average velocity. Equation 3 follows by eliminating t, by substitution, or from work and energy.
• Choose the equation by finding the missing quantity among s, u, v, a, t.
• Use a clear sign convention (for example, upward positive and a = −9.8 m/s² for vertical motion) and SI units.
• Starting from rest, v ∝ t, s ∝ t² and distances in successive equal times are in the ratio 1 : 3 : 5 : 7 : …

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Sources and Further Reading

This article is original EDUSAMBAM writing. The equations, derivations, constants and historical points were checked against the sources below. Diagrams are original and were not copied from any source.

• OpenStax, University Physics Volume 1: motion along a straight line, with derivations for constant acceleration, including the calculus approach. OpenStax — University Physics Volume 1.
• OpenStax, College Physics 2e: kinematics, motion equations and falling objects. OpenStax — College Physics 2e.
• NIST: standard acceleration of gravity, 9.80665 m/s². NIST — Fundamental Physical Constants.
• Encyclopaedia Britannica: Galileo Galilei (the law of falling bodies and Two New Sciences); Nicole Oresme (graphical treatment of uniformly changing motion).

Note on the numbers: numerical examples use rounded values and g = 9.8 m/s². Please check the derivations and the numbers against your own syllabus or exam board specification before relying on them.