Class 11 NCERT PHYSICS • Chapter 3: Motion in a Plane • Question 9

QUESTION 9

Displacement ( $\Delta \vec{r}$ ): The vector difference between the final and initial position vectors, $\Delta \vec{r} = \vec{r}_f - \vec{r}_i$ .
Average Velocity ( $\vec{v}_{\text{avg}}$ ): The ratio of net displacement to the total time interval, $\vec{v}_{\text{avg}} = \frac{\Delta \vec{r}}{\Delta t}$ .
Average Speed ( $v_{\text{avg}}$ ): The ratio of the total distance traveled to the total time interval, $v_{\text{avg}} = \frac{d}{\Delta t}$ .

Radius of the circular park, $R = 1\,\text{km}$
Total time taken for the round trip, $\Delta t = 10\,\text{min} = \frac{1}{6}\,\text{h}$
Path: $O \rightarrow P$ (radius), $P \rightarrow Q$ (quarter-circle arc), $Q \rightarrow O$ (radius)

Net Displacement: The cyclist starts at the center $O$ and returns to the center $O$ at the end of the trip. Since the initial and final positions are identical:

\Delta \vec{r} = 0\,\text{km}

Average Velocity: Average velocity is the net displacement divided by the total time:

\vec{v}_{\text{avg}} = \frac{0\,\text{km}}{10\,\text{min}} = 0\,\text{km} \cdot \text{h}^{-1}

Average Speed: First, we determine the total distance $d$ traveled. The path consists of two radii and a quarter-circle arc:

\begin{aligned} d &= OP + \text{arc } PQ + QO \\ &= R + \frac{1}{4}(2\pi R) + R \\ &= 2R + \frac{\pi R}{2} \end{aligned}

Substituting $R = 1\,\text{km}$ :

d = 2(1) + \frac{\pi(1)}{2} = 2 + \frac{\pi}{2}\,\text{km}

The average speed is the total distance divided by the total time $\Delta t = \frac{1}{6}\,\text{h}$ :

\begin{aligned} v_{\text{avg}} &= \frac{d}{\Delta t} \\ &= \frac{2 + \pi/2}{1/6} \\ &= 6 \left( 2 + \frac{\pi}{2} \right) \\ &= 12 + 3\pi \\ &\approx 21.42\,\text{km} \cdot \text{h}^{-1} \end{aligned}

(a) Net displacement:

0\,\text{km}

(b) Average velocity:

0\,\text{km} \cdot \text{h}^{-1}

21.42\,\text{km} \cdot \text{h}^{-1}