Module 3: Kinematics · Theory · Prerequisite: K6
The simplest motion there is, and the place where two habits are worth building: rejecting answers, and changing frames.
Mahavakya
x = x0 + vt is not a law of motion. It is the mathematical signature of one condition: that the acceleration is zero.
On this page
1. One condition, three descriptions
Uniform motion means constant velocity. That is the whole definition, and everything else follows from it.
Since v = dx/dt and v is constant, integrating gives
x = x0 + vt
with x0 the constant of integration, which is the position at t = 0. Nothing was assumed beyond the definition.
The same fact now has three faces, and it is worth seeing that they are one fact:
Algebraically — x = x0 + vt, a linear function of time.
Graphically — a straight line on the x–t graph, whose slope is v and whose intercept is x0. Constant slope means no curvature, so a = 0.
By area — on the v–t graph the motion is a horizontal line, and the area beneath it over a time t is the rectangle vt. Which is the displacement.
K6 built the machinery for the last two. This is the simplest case in which all three agree, and it is worth checking that they do, because from K8 onward you will use whichever is convenient without re-deriving it.
2. Equal displacements, not equal distances
Uniform motion is often described as covering equal distances in equal times. That is not quite right, and the imprecision matters.
Consider a body that travels 5 m forward each second for three seconds, then 5 m backward each second for three seconds. It covers equal distances in equal intervals throughout — and its velocity is certainly not constant, since it reversed.
Uniform motion means equal displacements in equal times. Displacement carries direction; distance does not. That single word is what excludes the reversal.
This is K1’s distinction doing real work. It is also a good example of how a nearly-correct statement can survive for years without being noticed, because the cases that expose it are not the ones usually asked.
3. Meeting problems
Two bodies in uniform motion along the same line. When and where do they meet?
The condition is simple: they meet when they are at the same place at the same time.
xA(t) = xB(t)
Write both position functions, set them equal, solve for t. That is the whole method, and it covers approach, pursuit and overtaking without needing a separate formula for each.
Worked example
A cyclist is 30 m ahead of a runner, travelling at 4 m/s. The runner sets off at 10 m/s in the same direction. When and where does the runner catch up?
Take the runner’s starting point as the origin, positive in the direction of travel:
xA = 30 + 4t xB = 10t
30 + 4t = 10t → 6t = 30 → t = 5 s
x = 10(5) = 50 m from the runner’s start
Notice the 6 in the working. That is 10 − 4, the rate at which the gap closes — and it appeared without being introduced. Section 5 gives it a name.
4. When the answer must be rejected
Now change the problem slightly. A body starts at x = 5 m moving at 3 m/s. A second starts at x = 20 m moving at 6 m/s, in the same direction. When do they meet?
5 + 3t = 20 + 6t
−15 = 3t → t = −5 s
The algebra is faultless and the answer is negative. What does that mean?
It means the two lines do intersect — five seconds before the clock started. Extended backwards, the second body was behind the first, and it has been pulling ahead ever since. For any t ≥ 0 they never meet, and they never will, because the one in front is also the faster.
A correct equation can produce an inadmissible answer. The mathematics knows nothing about when your clock was started, and it will happily report an intersection that lies outside the physical situation. The check is yours to make, and it belongs at the end of every problem — is the time positive, is the position on the part of the track that exists, is the root the one the question meant.
You will meet this again wherever a quadratic appears, and there it will produce two roots with one to discard. This is the same discipline in its simplest setting.
5. The same event from another frame
K1 said that motion is described relative to a frame someone chose, and that the choice is free. Here is the first case where making a different choice is useful.
Return to the runner and the cyclist. Instead of watching from the ground, ride with the cyclist. From there, the cyclist is stationary — and the runner approaches at 10 − 4 = 6 m/s, starting 30 m behind.
t = 30 / 6 = 5 s
One line of arithmetic, and the same answer. The two-body problem has become a one-body problem, because in the cyclist’s frame there is only one thing moving.
Two lines crossing, or one line reaching zero. The same instant, t = 5 s.
The figure shows what changed. In the ground frame the meeting is an intersection of two lines. In the cyclist’s frame it is a single line arriving at zero — and zero is where the cyclist is, permanently, because that is how the frame was chosen.
Nothing physical differs between the two panels. The bodies meet at one instant and the instant is the same. What changed is which quantities are simple — and choosing a frame is choosing what to make simple, exactly as choosing a graph was in K6.
6. Where relative velocity comes from
The 6 m/s used above deserves to be derived rather than assumed, because it is the beginning of a topic that runs through the rest of the module.
Define the position of B relative to A as the difference:
xBA = xB − xA
Differentiate: vBA = vB − vA
That is all. Relative velocity is not a new physical idea and not a formula to memorise — it is the derivative of a difference of positions, which is a difference of derivatives. Differentiate once more and you have the relative acceleration, aBA = aB − aA, which will matter a great deal later.
The subscript convention is worth fixing now, because it is the commonest source of sign errors in the topic: vBA means the velocity of B as measured by A — the first subscript is the thing observed, the second is the observer. Reverse them and you reverse the sign.
For the runner and cyclist: vBA = 10 − 4 = +6 m/s, so the runner approaches at 6 m/s in the positive direction, which is what the figure shows.
7. The boundary of the formula
x = x0 + vt is the most reliable equation in kinematics, and it applies to almost nothing.
It requires the velocity to be constant — which means no speeding up, no slowing down, and no change of direction, for the entire interval it is applied to. A body that is uniform for part of a journey and not for the rest needs the formula applied piecewise, or not at all.
So the equation is the signature of a condition rather than a law of nature. Where the condition holds, it holds exactly. Where it does not, no amount of care in applying it will help.
The habit worth building: before writing it down, ask whether the acceleration really is zero over the whole interval — and if the problem does not say, ask what would make it so.
One last observation from the x–t graph, since it catches students out. Two straight lines of very different steepness both represent uniform motion. The steeper one is faster, and neither is accelerating. Steepness is velocity; curvature is acceleration; a straight line has none.
Contemplation
Section 4 produced an answer that was correct and had to be thrown away. That is worth sitting with, because it is not what students are trained to expect. Arithmetic is checked; a correct calculation is supposed to be the end of the matter.
But an equation describes a model, and a model has a domain. The equation xA = xB knows nothing about when the clock was started or whether the road extends that far. It answers the question it was given, which is a slightly different question from the one you asked.
So the last step of a solution is not arithmetic. It is asking whether the answer belongs to the situation — and that step cannot be automated, because it requires knowing what the situation was.
Common misconceptions
1. Uniform motion means equal distances in equal times
Equal displacements. A body reversing regularly covers equal distances and is not in uniform motion.
2. A steeper x–t line means greater acceleration
It means greater velocity. Both lines are straight, so both bodies have zero acceleration.
3. A negative time means the calculation went wrong
Usually it means the intersection lies before t = 0. The calculation is right; the answer is outside the situation and must be rejected.
4. Relative velocity is a separate formula
It is the derivative of a difference of positions. If you can differentiate, you can reconstruct it in one line.
5. vAB and vBA are interchangeable
They differ by a sign. The first subscript is what is observed; the second is who is watching.
Reflection
- Two bodies in uniform motion along the same line. Under what conditions do they never meet for any t, positive or negative?
- In the cyclist’s frame, what does the ground do? Draw its position–time graph.
- You are told vBA = 0. What does that say about the two bodies, and what does it not say?
Key takeaways
- Uniform motion is constant velocity; x = x0 + vt follows by integration.
- Equal displacements in equal times, not equal distances.
- Two bodies meet when xA(t) = xB(t). One condition covers approach, pursuit and overtaking.
- A correct equation can give an inadmissible answer. Check the time is positive and the position is real.
- Working in one body’s frame turns a two-body problem into a one-body problem.
- vBA = vB − vA, derived by differentiating a difference of positions.
- The formula is the signature of a = 0 and applies only where that holds.
What comes next
Zero acceleration gave a straight line and a single equation. Constant acceleration — the next simplest case — gives a parabola and four equations, all of which turn out to be the same statement written four ways. K8 derives them, and then applies them to the one acceleration everyone has already met.
Prerequisite: K6 · Reading: 14 min · Practice: 25 min · Difficulty: Beginner to Intermediate