A Round Trip Can Have Zero Velocity but a Real Speed

An auto-rickshaw takes you $3$ km to the market and $3$ km back in $30$ minutes. How fast did you go? $12$ km/h if you count the ground covered — but zero if you only ask how far you ended up from home.

Both answers are correct; they measure different things. Straight-line motion needs precise ideas of displacement, velocity and acceleration, and graphs to picture them.

This part covers average and instantaneous quantities, derivatives, reading graphs, and uniform versus non-uniform motion.

Average velocity is total displacement divided by total time, while average speed is total path length divided by total time; the instantaneous values are the limits of these averages as the time interval shrinks to zero.

$$\bar{v}=\frac{\Delta x}{\Delta t}\text{average speed}=\frac{\text{path length}}{\Delta t}v=\underset{\Delta t\to 0}{\lim }\frac{\Delta x}{\Delta t}$$

Worked example. A student walks $400$ m east in $5$ min, then $300$ m west in $3$ min.

- Displacement $=400-300=100$ m east; path length $=700$ m
- Time $=8$ min $=480$ s

$$\bar{v}=\frac{100}{480}\approx 0.21\text{ m/s east}\text{average speed}=\frac{700}{480}\approx 1.46\text{ m/s}$$

An everyday example. The auto-rickshaw round trip has average velocity $0$ but average speed $12$ km/h.

The substance. Instantaneous speed always equals the magnitude of instantaneous velocity, but average speed is usually larger than the magnitude of average velocity.

**Instantaneous velocity is the rate of change of position, $v=\frac{dx}{dt}$, and instantaneous acceleration is the rate of change of velocity, $a=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$.

Worked example 1.** $x=5+3t+2t^2$ (metres, seconds).

$$v=\frac{dx}{dt}=3+4tv(2)=11\text{ m/s}a=\frac{dv}{dt}=4{\text{ m/s}}^2$$

For comparison, $x(0)=5$ m and $x(2)=19$ m, so the average velocity over the first $2$ s is $\frac{14}{2}=7$ m/s.

Worked example 2 — a reversal. $x=t^3-6t^2+9t$.

$$v=3t^2-12t+9=3(t-1)(t-3)a=6t-12$$

Velocity is zero at $t=1$ s and $t=3$ s. Since $x(1)=4$ m and $x(3)=0$, the body moves $4$ m forward, then $4$ m back: **distance $8$ m, displacement $0$ in the first $3$ s.

An everyday example. A car's speedometer** shows instantaneous speed, the magnitude of $\frac{dx}{dt}$ at that moment.

The substance. Negative acceleration does not always mean slowing down — a body slows down only when $a$ and $v$ have opposite signs.

The slope of an x-t graph is velocity, the slope of a v-t graph is acceleration, and the area under a v-t graph is displacement.

Worked example 1 — x-t slope. A straight x-t line through $(0\text{ s},10\text{ m})$ and $(5\text{ s},40\text{ m})$ gives $v=\frac{40-10}{5}=6$ m/s.

Worked example 2 — a v-t trapezium. A car starts from rest, reaches $20$ m/s in $10$ s, keeps that speed for $20$ s, then stops in $5$ s.

- Accelerations (slopes): $\frac{20}{10}=2$ m/s${}^2$, then $0$, then $-\frac{20}{5}=-4$ m/s${}^2$
- Displacement (area):

$$\frac{1}{2}(10)(20)+(20)(20)+\frac{1}{2}(5)(20)=100+400+50=550\text{ m}$$

$$\text{average velocity}=\frac{550}{35}\approx 15.7\text{ m/s}$$

Area below the time axis counts as negative displacement; for distance, add all areas as positive.

An everyday example. A metro train between two stations speeds up, cruises and brakes, giving exactly this trapezium-shaped v-t graph.

The substance. The area under an x-t graph has no physical meaning — only v-t areas give displacement.

Uniform motion covers equal displacements in equal time intervals, giving a straight x-t line and a flat v-t line; any other motion is non-uniform, with a curved x-t graph.

The graph shapes:

- At rest — x-t horizontal line; v-t lies on the time axis
- Uniform velocity — x-t straight sloping line; v-t horizontal line
- Uniform acceleration — x-t parabola curving upward; v-t straight rising line
- Uniform retardation to rest — x-t curve that flattens out; v-t straight falling line

Worked example 1 — uniform. A train at a steady $72$ km/h $=20$ m/s has $x=20t$, so in $30$ s it covers $600$ m, and its v-t graph is flat at $20$ m/s.

Worked example 2 — non-uniform. A stone dropped from rest has $v=9.8t$ and $x=4.9t^2$. At $t=2$ s, $v=19.6$ m/s and $x=19.6$ m; its x-t graph is a parabola.

An everyday example. A train on a clear straight track can move uniformly, while a bus in city traffic is always non-uniform.

The substance. Real x-t graphs can never show two positions at one instant, and real v-t graphs cannot have vertical jumps, which would need infinite acceleration.