Q.A man walks on a straight road from his home to a market $2.5\ \text{km}$ away with a speed of $5\ \text{km h}^{-1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5\ \text{km h}^{-1}$. What is the

Concept: Instantaneous Velocity — but here we need average velocity (net displacement / time) and average speed (total distance / time). The key is to track the man's position at each time.

Step 1: Find times for each leg.

Home to market: $t_1 = \frac{2.5}{5} = 0.5\ \text{h} = 30\ \text{min}$.

Market to home: $t_2 = \frac{2.5}{7.5} = \frac{1}{3}\ \text{h} = 20\ \text{min}$.

Total round trip time = $50\ \text{min}$.

Step 2: For each time interval, find net displacement and total distance.

• (i) 0 to 30 min: Man reaches market. Displacement = $+2.5\ \text{km}$, distance = $2.5\ \text{km}$.

  Average velocity = $\frac{2.5}{0.5} = 5\ \text{km h}^{-1}$.

  Average speed = $\frac{2.5}{0.5} = 5\ \text{km h}^{-1}$.

• (ii) 0 to 50 min: Man returns home. Displacement = $0$, distance = $5\ \text{km}$.

  Average velocity = $0$.

  Average speed = $\frac{5}{50/60} = 6\ \text{km h}^{-1}$. …

The core idea: displacement vs. distance

This is a classic trap. Average velocity is net displacement divided by total time — displacement is a vector, so it only cares about the start and end points, not the path taken. Average speed is total path length divided by total time — it cares about every step walked. When the man turns around and comes back, his displacement can shrink or even vanish even though he has walked a long way; that's exactly why the question warns against telling the tired man his average speed was zero.

First, find the time for each leg.

Home to market: distance $=2.5\ \text{km}$, speed $=5\ \text{km/h}$, so time $=\dfrac{2.5}{5}=0.5\ \text{h}=30\ \text{min}$.

Market back to home: distance $=2.5\ \text{km}$, speed $=7.5\ \text{km/h}$, so time $=\dfrac{2.5}{7.5}=\dfrac{1}{3}\ \text{h}=20\ \text{min}$.

So the round trip takes $30+20=50\ \text{min}$ in total.

(a) Magnitude of average velocity

Average velocity $=\dfrac{\text{net displacement}}{\text{total time}}$; we report its magnitude.

1. 0 to 30 min: At $t=30\ \text{min}$ he has just reached the market — displacement from home $=2.5\ \text{km}$, time $=0.5\ \text{h}$.

   $v_{\text{avg}} = \frac{2.5}{0.5} = 5\ \text{km/h}$

2. 0 to 50 min: At $t=50\ \text{min}$ he is back home, so net displacement $=0$.

   $v_{\text{avg}} = \frac{0}{5/6} = 0$

3. 0 to 40 min: He spent the first 30 min walking to the market, so by $t=40\ \text{min}$ he has been returning for $40-30=10\ \text{min}=\dfrac{1}{6}\ \text{h}$. Distance covered on the return leg $=7.5\times\dfrac16=1.25\ \text{km}$. His displacement from home $=2.5-1.25=1.25\ \text{km}$, over a total time of $\dfrac{2}{3}\ \text{h}$: $v_{\text{avg}} = \frac{1.25}{2/3} = 1.875\ \text{km/h}$ …

Concept: One Pair of Piecewise Functions, Evaluated at Every Requested Time

Method: Build $x(t)$ and $s(t)$ Once, Then Substitute (instead of re-reasoning for each part)

Rather than working out the average velocity and average speed separately for each of the three time windows using fresh leg-by-leg reasoning, this method writes down the man's position function $x(t)$ and his cumulative-distance-walked function $s(t)$ once, as explicit piecewise formulas — then every part of the question becomes a simple substitution.

Let $t$ be in minutes, $x=0$ at home, market at $x=2.5\ \text{km}$.

Steps

1. Outbound leg ($5\ \text{km h}^{-1} = \tfrac{1}{12}\ \text{km/min}$), valid for $0\le t\le30$:

Check: $x(30) = 30/12 = 2.5\ \text{km}$. ✓.

2. Return leg ($7.5\ \text{km h}^{-1} = 0.125\ \text{km/min}$), valid for $30\le t\le50$:

($x$ counts back down toward zero; $s$ keeps accumulating the extra distance walked on the way back.) Check: $x(50)=2.5-0.125(20)=0$ ✓, $s(50)=2.5+2.5=5\ \text{km}$ (total round-trip distance) ✓.

3. **(a)(i) $t=30$: ** $|v_{\text{avg}}| = \dfrac{|x(30)|}{30\,\text{min}} = \dfrac{2.5}{0.5\,\text{h}} = 5\ \text{km h}^{-1}$

4. **(a)(ii) $t=50$: ** $x(50)=0$, so $|v_{\text{avg}}| = \dfrac{0}{50/60\,\text{h}} = 0$

5. **(a)(iii) $t=40$: ** $x(40) = 2.5-0.125(10) = 1.25\ \text{km}$, so $|v_{\text{avg}}| = \dfrac{1.25}{40/60\,\text{h}} = \dfrac{1.25}{2/3} = 1.875\ \text{km h}^{-1}$

6. **(b)(i) $t=30$: ** $s(30) = x(30) = 2.5\ \text{km}$, so speed $= \dfrac{2.5}{0.5} = 5\ \text{km h}^{-1}$ …