Module 3: Kinematics · Theory · Prerequisites: K7, Foundation 9
Four equations that every student memorises, and one assumption that produces all of them.
Mahavakya
The equations of motion are not four facts. They are four shadows cast by one physical assumption.
On this page
1. Deriving them
Assume the acceleration is constant. That is the only input; everything else is calculus.
First equation. Since a = dv/dt with a constant, integrate from the initial velocity u:
v = u + at
Second equation. Now v = ds/dt, so substitute and integrate again:
s = ut + ½at²
Third equation. Here time is not wanted, so use Foundation 9’s identity a = v dv/ds, separate, and integrate over position:
v² = u² + 2as
Fourth equation. Since the velocity changes at a constant rate, its average over the interval is the mean of its endpoints:
s = ½(u + v)t
K6 obtained the third equation a different way — from the area of a trapezium and the difference of two squares, with no calculus at all. And integrating the acceleration–time graph gives the first.
Three independent routes to the same results. That is the strongest evidence available that these are consequences of constant acceleration rather than laws to be stored — and it is why the fourth equation is not independent either, being the second and first combined.
Note also that the fourth equation is where K3’s warning was pointing. vavg = (u + v)/2 is not a general result about averages; it holds precisely because the acceleration is constant, and nowhere else.
2. Why t² appears
A small question worth answering, because students use the term for years without asking.
Under uniform motion the velocity is a constant, and integrating a constant gives a term in t. That is K7’s x = x0 + vt.
Under constant acceleration the velocity is u + at — it contains t itself. Integrating t gives t²/2, and there is the ½at².
So the quadratic dependence is not a feature of the formula. It is what one extra layer of change produces, and it is why K6’s degree rule works: each integration raises the power by one, because that is what integration does.
3. Distance in the nth second
Examinations ask for the distance covered during a particular second — the fifth, say, meaning the interval from t = 4 to t = 5. There is a standard formula for this. You do not need it.
Worked example
A body starts at 6 m/s with a constant acceleration of 2 m/s². Find the distance covered during the fifth second.
By subtraction. Positions at t = 5 and t = 4:
s(5) = 6(5) + ½(2)(25) = 55 s(4) = 6(4) + ½(2)(16) = 40
difference = 15 m
By average velocity. Over one second, distance is the average velocity across it:
v(4) = 6 + 8 = 14 v(5) = 6 + 10 = 16 mean = 15 m/s
15 × 1 s = 15 m
Two routes, one line each, no formula. The second uses the fourth equation over an interval of one second, which is why it is so short.
The language is the harder part. “During the fifth second” means 4 ≤ t ≤ 5, not the first five seconds. That misreading costs more marks than the arithmetic ever does.
4. Free fall is not a new topic
Near the Earth’s surface, and ignoring air resistance, every body falls with the same constant acceleration of magnitude g ≈ 9.8 m/s².
Constant acceleration. So the four equations apply unchanged, and free fall is not a new chapter — it is the previous three sections with one particular value substituted.
Take upward as positive. Then a = −g throughout — while rising, at the top, and while falling. The acceleration never changes and never vanishes.
Take downward as positive instead and a = +g, with every velocity and displacement flipping sign to match. Both conventions are correct. What matters is choosing before you write, and not changing halfway.
K1 said the sign convention is a decision rather than a fact. Free fall is where that stops being an abstract remark, because g is a magnitude — a positive number — and the sign belongs to the coordinate system, not to gravity.
5. Symmetry, and reading it off the coefficients
A body projected upward at u reaches its highest point when v = 0, so from the first equation tup = u/g. Setting the displacement to zero in the second gives a total flight time of 2u/g, exactly twice as long. Rise and fall take equal times, and the speed at any height on the way down equals the speed at that height on the way up.
All of which can be got more elegantly. Ask when the body is at some height h:
h = ut − ½gt²
½gt² − ut + h = 0
A quadratic in t, with two roots — the body passes every reachable height twice. And rather than solving it, read the relations between the roots straight off the coefficients:
t1 + t2 = 2u/g
t1t2 = 2h/g
The sum is the total time of flight, so the two crossings are symmetric about the midpoint — which is the symmetry, obtained without computing anything.
Sum of roots = 2u/g. Product of roots = 2h/g.
Worked example
A ball thrown vertically upward passes a window at t = 2 s and again at t = 8 s. Taking g = 10 m/s², find the launch speed, the window’s height, and the maximum height reached.
From the sum of the roots: 2 + 8 = 2u/g, so u = 50 m/s.
From the product: (2)(8) = 2h/g, so h = 80 m.
Maximum height: u²/2g = 2500/20 = 125 m.
No quadratic was solved. Two facts about the roots were read from the coefficients, and the third result followed from the standard equations. The symmetry of the flight turns out to be encoded in the structure of the quadratic.
6. Three traps
Acceleration is not zero at the top. The velocity is momentarily zero; the acceleration is −g, as it is throughout. If it were zero the ball would remain there. K4 made this point in general; here is where it bites.
Release from a moving platform. A stone dropped from a balloon rising at 5 m/s does not start from rest — it starts at 5 m/s upward, rises a little further, and only then falls. Setting u = 0 is the commonest error in this topic.
Displacement is not distance. A ball thrown up from a cliff of height H and landing at the base has displacement −H. The distance travelled is 2Hmax + H, because it went up first. Solving y = −H gives the time correctly and the distance not at all.
7. Where the equations stop working
Every one of the four equations assumed constant acceleration, and each fails the moment that assumption does.
Non-constant acceleration. If a depends on time, position or velocity, these equations do not apply at all — not even with an average value substituted. K9 handles that case, and it needs the definitions rather than the results.
Reversal within the interval. s = ut + ½at² gives displacement. If the body turns around during the interval, the distance is larger, and finding it means splitting at the turning point first.
Unsigned quantities. Substituting magnitudes and hoping the signs sort themselves out fails immediately in free fall, where u and a routinely have opposite signs.
A check you can run on your own work. Given a position function, differentiate it twice and read off x0, u and a; then rebuild x = x0 + ut + ½at² and see whether it reproduces what you started with.
For x = 5 + 10t + 3t²: v = 10 + 6t and a = 6, so the model predicts 5 + 10t + 3t². It agrees, and the motion is confirmed to be uniformly accelerated — with no answer key involved.
Contemplation
Most students meet these four equations as a list to be memorised, and they are not four things. Start from a = constant; integrate once and you have the first; integrate again and you have the second; eliminate time and you have the third; average the velocity and you have the fourth.
One assumption, four shadows. And the difference between the student who has memorised them and the student who has seen this is not the exam where the equations apply — both do fine there. It is the exam where they do not, and only one of the two notices.
That is the real reason to derive things. Not elegance, and not rigour for its own sake — but that a result you can reconstruct carries its conditions with it, and a result you have merely stored does not.
Common misconceptions
1. Acceleration is zero at the highest point
The velocity is zero for an instant. The acceleration is −g the whole time, which is what turns the motion around.
2. A body released from a moving platform starts from rest
It keeps whatever velocity the platform had. Dropped from a rising balloon, it goes up before it comes down.
3. g is negative
g is a magnitude, about 9.8 m/s². The sign of the acceleration comes from your coordinate choice: −g with up positive, +g with down positive.
4. vavg = (u + v)/2 always
Only under constant acceleration. K3 flagged this; here is the condition that makes it true.
5. Heavier bodies fall faster
Not in free fall, where the mass appears nowhere in any of the four equations. What makes a feather fall slowly is air resistance, which is precisely what free fall neglects.
Reflection
- The quadratic for height has two roots. What happens to them as h approaches the maximum height, and what does that mean physically?
- A stone is dropped from a balloon descending at 5 m/s. How does the answer differ from the rising case?
- Which of the four equations would you keep if you could keep only one, and why?
Key takeaways
- All four equations follow from a = constant by integration. Three independent routes reach the same results.
- The t² appears because the velocity itself contains t.
- Distance in the nth second: subtract two positions, or average the velocities across that second. No formula needed.
- Free fall is constant acceleration with a = −g if up is positive. g is a magnitude; the sign is yours.
- Setting the height to h gives a quadratic whose roots sum to 2u/g and multiply to 2h/g.
- Acceleration is never zero during free fall, including at the top.
- The equations fail for non-constant acceleration, and give displacement rather than distance when the body reverses.
What comes next
Section 7 listed the conditions under which these equations hold, and the first of them — constant acceleration — is the one most real motions violate. A parachutist, a stone in water, a mass on a spring: in each the acceleration depends on time, or position, or velocity. K9 takes those three cases, and finds that each needs a different mathematical move.
Prerequisites: K7 and Foundation 9 · Reading: 18 min · Practice: 40 min · Difficulty: Intermediate