Q.Explain clearly, with examples, the distinction between:

Key Idea: Instantaneous Velocity vs. Path-Dependent Quantities

(a) Magnitude of Displacement vs. Total Path Length

• Displacement is the straight-line distance between initial and final positions. Its magnitude is $|x_f - x_i|$.
• Total path length is the actual distance travelled along the path, which can be longer if the particle reverses direction.

Example: A particle moves from $x=0$ to $x=5$ m, then back to $x=2$ m.

Magnitude of displacement = $|2-0| = 2$ m.

Total path length = $5 + 3 = 8$ m.

Clearly, $8 \ge 2$.

(b) Magnitude of Average Velocity vs. Average Speed

• Average velocity = $\frac{\text{displacement}}{\text{time interval}}$, so its magnitude is $\frac{|\text{displacement}|}{\Delta t}$.
• Average speed = $\frac{\text{total path length}}{\Delta t}$. …

Why This Distinction Matters

Think of a particle moving along a line. If it goes from A to B and then back to A, its displacement is zero — it ended where it started. But the path length is the total ground it covered, which is clearly not zero. This simple example shows that these two quantities are fundamentally different.

The same idea carries over to velocity and speed. Average velocity depends only on the net displacement (where you ended vs where you started), while average speed depends on the total path length (how much road you actually travelled). One is a vector, the other a scalar — and they are not the same thing.

(a) Magnitude of Displacement vs Total Path Length

1. Definitions in one dimension

Let a particle move along the $x$-axis. Suppose at time $t_1$ it is at position $x_1$, and at time $t_2$ it is at $x_2$.

• Displacement: $\Delta x = x_2 - x_1$. Its magnitude is $|\Delta x|$.
• Total path length: The sum of all distances travelled, regardless of direction. If the particle moves back and forth, you add up every segment.

2. Why path length ≥ |displacement|

Imagine the particle starts at $x=0$, goes to $x=+5$, then back to $x=+2$.

Displacement = $+2 - 0 = 2$, so $|\Delta x| = 2$.

Path length = $5 + 3 = 8$. Clearly $8 \geq 2$.

In general, the shortest distance between two points is the straight line. Any back-and-forth motion adds extra distance. So:

3. When does equality hold?

Equality happens when the particle never reverses direction — it moves monotonically from start to finish. In one dimension, that means the velocity has the same sign throughout the interval.

(b) Magnitude of Average Velocity vs Average Speed

1. Definitions

• Average velocity: $\vec{v}_{\text{avg}} = \frac{\Delta x}{\Delta t}$. Its magnitude is $|\vec{v}_{\text{avg}}| = \frac{|\Delta x|}{\Delta t}$.
• Average speed: $s_{\text{avg}} = \frac{\text{Total path length}}{\Delta t}$.

2. Why average speed ≥ |average velocity|

From part (a), we already have:

Dividing both sides by the positive time interval $\Delta t$: …

Concept: A General Proof via the Triangle Inequality

Method: Triangle-Inequality Proof over a Fine Partition (holds for any motion, not just one example)

A single back-and-forth example (out and partway back) makes the claim plausible, but doesn't prove it for every possible motion. This method gives a general proof, valid for any one-dimensional motion, by partitioning the time interval into many small pieces and applying the triangle inequality for sums of real numbers.

Steps — part (a): path length $\ge$ |displacement|

1. Partition the time interval $[t_1, t_2]$ into $N$ small sub-intervals, each short enough that the particle doesn't reverse direction within any single sub-interval. Let $\Delta x_i$ be the (signed) displacement during the $i$-th sub-interval.

2. Total displacement is the sum of the pieces — this is just telescoping (each interior boundary cancels, leaving only the start and end position):

3. Total path length is the sum of the magnitudes of the pieces (since within each tiny sub-interval, distance travelled = $|\Delta x_i|$, as there's no reversal inside it):

4. Apply the triangle inequality for sums, $\left|\sum_i \Delta x_i\right| \le \sum_i |\Delta x_i|$, which holds for any real numbers $\Delta x_i$ regardless of their signs:

5. Equality condition, straight from the triangle inequality itself: equality holds if and only if all the terms $\Delta x_i$ have the same sign — i.e., the particle never reverses direction over $[t_1,t_2]$.

Steps — part (b): average speed $\ge$ |average velocity|

6. Divide the inequality from Step 4 by the (positive) time interval $\Delta t = t_2-t_1$:

using the definitions $\vec v_{\text{avg}} = \Delta x/\Delta t$ and $s_{\text{avg}} = L/\Delta t$. …