Motion in a Straight Line

To describe motion along a straight line, first fix an origin and a positive direction. A car’s position $x$ is then just a signed number, its distance from the origin, with the sign telling you which side it is on. There is no motion to speak of until you also track how $x$ changes with time $t$.

Two different quantities describe “how far it moved”, and conflating them is the single most common error in this chapter. Path length (or distance) is the total length of the actual path traced out, always positive, and it only ever adds up. Displacement is the change in position, $\Delta x = x_2 - x_1$, a straight-line, signed quantity that only cares about where you started and where you ended up.

Say a student walks 4 m east from a lamppost and then 3 m back west. The path length is $4+3=7$ m, she physically walked seven metres of pavement. But her displacement is only $4-3=1$ m east, since that is the net change in her position. If instead she walked 4 m east and then 4 m back west, landing exactly where she started, the path length would still be 8 m, but the displacement would be zero. This is why displacement can be zero, or even negative, while path length never can: path length only cares about effort, displacement only cares about the net result.

Average velocity over an interval is displacement divided by the time taken:

Average speed, by contrast, is path length divided by the same time interval. Since path length and displacement can differ, so can these two: average speed is always $\ge$ the magnitude of average velocity, never less.

The round trip makes this vivid. Imagine someone jogs 500 m down a straight track and jogs back to the start, taking 250 s in total. Their displacement for the whole trip is zero, so their average velocity is zero, even though they were clearly moving the entire time. Their average speed, meanwhile, is the full 1000 m path length over 250 s, a solidly nonzero $4\text{ m/s}$. Average velocity answers “where did you net up, divided by how long it took”; average speed answers “how hard were you actually working”, and a round trip is the cleanest case where the two completely part ways.

Average velocity hides everything that happens inside the interval, a car could have been stopped for half the time and speeding for the rest, and the average would never show it. To capture the motion at one exact instant, shrink the interval $\Delta t$ down toward zero:

This is the instantaneous velocity, the slope of the tangent to the position-time graph at that instant (equivalently, its derivative). It is a signed quantity: positive while $x$ is increasing, negative while it is decreasing, and zero exactly where the $x\text{-}t$ graph momentarily flattens out, such as at the top of a ball’s vertical throw.

Instantaneous speed is simply $|v|$, the magnitude of this instantaneous velocity, and it is here, unlike for the averaged versions, that speed and velocity are guaranteed to match in size. The reason is almost definitional: over an infinitesimally short instant, the path traced out and the straight-line displacement become the same infinitesimal length, there is no room left for the path to double back on itself within a vanishing interval the way it can over a long one. This is worth remembering as a rule you can trust: speed equals $|v|$, the magnitude of the instantaneous velocity, and essentially never the magnitude of an average velocity over a finite stretch of time.

When velocity itself changes with time, that rate of change is acceleration. Just as velocity was defined from position, acceleration is defined from velocity in exactly the same two-step way, first an average, then an instantaneous limit:

Geometrically, $a$ is the slope of the $v\text{-}t$ graph, exactly as $v$ was the slope of the $x\text{-}t$ graph. Its SI unit is $\text{m/s}^2$.

A very common slip is to equate “negative acceleration” with “slowing down”, or “deceleration”. They are not the same thing, what matters is whether $a$ and $v$ point the same way or opposite ways, not the bare sign of $a$ on its own:

• If $a$ and $v$ have the same sign (both positive, or both negative), speed is increasing, regardless of whether $a$ itself is positive or negative.
• If $a$ and $v$ have opposite signs, speed is decreasing, this is the true meaning of deceleration.

Concretely: a car moving in the negative direction ($v<0$) that is pressed harder in the negative direction ($a<0$) is speeding up, even though its acceleration is negative. The same car braking, with $a>0$ opposing its negative velocity, is slowing down despite a positive $a$. Always check the relative signs, never the sign of $a$ in isolation.

When acceleration is constant, the $v\text{-}t$ graph is simply a straight line, and that one fact is enough to derive all three standard kinematic equations directly from its slope and the area underneath it. Let $u$ be the velocity at $t=0$, and $v$ the velocity at time $t$.

These three hold only while $a$ is truly constant over the interval in question, re-derive from scratch (or split into pieces of constant $a$) whenever it is not.

All velocities so far were measured against the ground. But motion is always relative to something, and often the more useful reference frame is another moving object. The velocity of object A relative to object B is defined as:

with the usual sign convention (one direction positive, the opposite negative) applied consistently to both $v_A$ and $v_B$.

Take two trains on parallel straight tracks. Train A moves at $+30\text{ m/s}$ and train B at $+20\text{ m/s}$, both taking the same direction as positive. To a passenger on B, train A appears to move forward at only $v_{AB} = 30-20 = 10$ m/s, much slower than it looks from the ground, since B is chasing it down at nearly the same speed. Now suppose B instead moves at $-20\text{ m/s}$, i.e. toward A. Then $v_{AB} = 30-(-20) = 50$ m/s: to B’s passenger, A now rushes past much faster than its ground speed, because the two are closing in on each other. Relative velocity is simply what one object’s motion looks like from inside the other’s reference frame.

A classic exercise question from this chapter, with an original worked solution.