Distance, velocity and acceleration

Distance vs displacement. Distance is the total length of the path a body actually travels — always positive, a scalar, with no notion of direction. Displacement is the shortest straight-line gap between where the body started and where it ended up, together with a direction — a vector, which can be positive, negative, or even zero if the body returns to its starting point. A runner who does one full lap of a 400 m track has covered a distance of 400 m but a displacement of 0 m.

Speed vs velocity. Speed is how much distance is covered per unit time:

where: distance is in metres and time is in seconds, giving speed in m/s. Velocity is the equivalent idea for displacement — speed with a direction attached:

Speed is a scalar, velocity is a vector, and both share the SI unit m/s. Uniform velocity means equal displacements occur in equal intervals of time, along a fixed direction; the two quantities have the same magnitude only when a body moves in a straight line without reversing or doubling back.

Acceleration and retardation. Acceleration is the rate of change of velocity with time — how quickly the velocity itself is speeding up, slowing down, or otherwise changing:

where: $u$ is the initial velocity, $v$ is the final velocity, $t$ is the time taken, and $a$ comes out in m/s². A positive value of $a$ (velocity rising) is ordinary acceleration; a negative value (velocity falling, so $v<u$) describes the same physical situation the syllabus calls retardation, or deceleration — no separate formula is needed, only a sign check. The figure below lines the five quantities up side by side, so the scalar/vector split and the "rate of a rate" relationship between velocity and acceleration are visible at a glance.

The goal of this derivation is to show exactly how acceleration, as the rate of change of velocity, turns into the working formula $a=(v-u)/t$, and how a single sign convention inside that same formula quietly covers both speeding up and slowing down without needing a second equation for retardation.

Step 1. Start from the definition. Acceleration is defined as the change in velocity divided by the time taken for that change to happen: $a = \dfrac{\text{change in velocity}}{\text{time taken}}$.

Step 2. Name the change in velocity precisely. If a body's velocity at the start of an interval is $u$ (initial velocity) and its velocity at the end of that interval is $v$ (final velocity), then the change in velocity is simply $v-u$. Substituting this into Step 1 gives $a = \dfrac{v-u}{t}$, the working formula, with $t$ the time taken for the velocity to go from $u$ to $v$.

Step 3. Read the sign. If $v>u$, the numerator $v-u$ is positive, so $a$ is positive — the body is genuinely accelerating, gaining speed in its direction of motion. If $v<u$, the numerator is negative, so $a$ comes out negative even though nothing about the formula has changed — this negative acceleration is exactly what the syllabus calls retardation or deceleration.

Step 4. Rearrange for use. The same relation can be flipped to find a final velocity when acceleration is known: $v = u + at$. This rearranged form is what makes motion-under-gravity questions solvable, since $g$ plays the role of $a$ once a body is falling or rising freely.

Question: A car travels 150 m in 10 s at uniform speed, then speeds up uniformly and reaches a velocity of 30 m/s after another 5 s. Find (a) its speed during the first stage, and (b) its acceleration during the second stage.

Step 1. Speed in stage 1 comes straight from the distance–time data: $\text{Speed} = \dfrac{\text{Distance}}{\text{Time}} = \dfrac{150\text{ m}}{10\text{ s}} = 15\text{ m/s}$.

Step 2. For stage 2, identify $u$, $v$ and $t$. The car enters this stage already moving at $15\text{ m/s}$ (the speed found in Step 1), so $u=15\text{ m/s}$; it leaves the stage at $v=30\text{ m/s}$; and the stage lasts $t=5\text{ s}$.

Step 3. Apply $a=(v-u)/t$: $a = \dfrac{30-15}{5} = \dfrac{15}{5} = 3\text{ m/s}^2$.

The car's speed in stage 1 is 15 m/s, and its acceleration in stage 2 is 3 m/s², meaning its velocity rises by 3 m/s every second.

Competitive stretch

A stone is dropped from rest and falls freely under gravity, $g=9.8\text{ m/s}^2$. Find its velocity after 3 s, and state whether this is acceleration or retardation. Here $u=0\text{ m/s}$ (dropped from rest) and $a=g=9.8\text{ m/s}^2$ downward, in the same direction as the motion. Using $v=u+at$: $v = 0 + (9.8)(3) = 29.4\text{ m/s}$. Since velocity is increasing (0 to 29.4 m/s) in the direction of motion, this is acceleration, not retardation — had the stone instead been thrown straight upward, it would slow down as it rises, which would be retardation even though the magnitude of $g$ used is identical.

• Confusing distance and displacement in a numerical: always check whether the body returns to or passes through its starting point — if so, displacement is smaller than distance, possibly zero, while distance keeps growing.
• Treating "speed" and "velocity" as always interchangeable: they share a magnitude only for straight-line motion with no reversal; the moment direction is mentioned, use velocity, not speed.
• Inventing a separate formula for retardation: there isn't one — $a=(v-u)/t$ already covers it, and a negative result is the signal that the body is slowing down.
• Assuming acceleration only means "getting faster": a body slowing down, or one whose velocity reverses direction, is also accelerating in the physics sense, because acceleration tracks any rate of change of velocity.

Ask the tutor: Still mixing up speed with velocity, or want more practice turning a motion-under-gravity story problem into $u$, $v$, $t$ and $a$? Ask the AI tutor for a fresh set of problems on this topic.