NCERT Solutions

We need to calculate the average speed and average velocity magnitude for a taxi trip given distance, displacement, and time.

Concept: Average Speed

The average speed is the total distance traveled divided by the total time taken. The average velocity magnitude is the net displacement divided by the total time taken.

$$v_{\text{avg}} = \frac{\text{Total Distance}}{\Delta t}, \quad |\vec{v}_{\text{avg}}| = \frac{|\text{Displacement}|}{\Delta t}$$

Given:

• Total distance traveled: $d = 23\,\text{km}$
• Displacement (straight-line distance): $\Delta x = 10\,\text{km}$
• Time taken: $\Delta t = 28\,\text{min}$

Solving:

1. Convert time from minutes to hours

$$\begin{aligned} \Delta t &= \frac{28}{60}\,\text{h} \\ &= \frac{7}{15}\,\text{h} \end{aligned}$$

2. Calculate average speed

$$\begin{aligned} \text{Average speed} &= \frac{d}{\Delta t} \\ &= \frac{23\,\text{km}}{7/15\,\text{h}} \\ &= \frac{23 \times 15}{7}\,\text{km}\cdot\text{h}^{-1} \\ &= \frac{345}{7}\,\text{km}\cdot\text{h}^{-1} \\ &\approx 49.29\,\text{km}\cdot\text{h}^{-1} \end{aligned}$$

3. Calculate magnitude of average velocity

$$\begin{aligned} |\vec{v}_{\text{avg}}| &= \frac{\Delta x}{\Delta t} \\ &= \frac{10\,\text{km}}{7/15\,\text{h}} \\ &= \frac{10 \times 15}{7}\,\text{km}\cdot\text{h}^{-1} \\ &= \frac{150}{7}\,\text{km}\cdot\text{h}^{-1} \\ &\approx 21.43\,\text{km}\cdot\text{h}^{-1} \end{aligned}$$

4. Comparison Since the path is circuitous, the distance ($23\,\text{km}$) is greater than the displacement ($10\,\text{km}$). Therefore, the average speed is not equal to the magnitude of the average velocity.

Answer:

(a) Average speed:

$$49.29\,\text{km}\cdot\text{h}^{-1}$$

(b) Magnitude of average velocity:

$$21.43\,\text{km}\cdot\text{h}^{-1}$$

No, the two are not equal.