An athlete runs along a circular track of circumference 400 m with a uniform speed 15 m/s . What is the change in velocity when the runs through a distance (i) 100 m (ii) 200 m ?

Explanation

To solve the problem step by step, we will analyze the athlete's movement along the circular track and calculate the change in velocity for both distances (100 m and 200 m).

### Step 1: Understanding the Circular Track
The athlete runs along a circular track with a circumference of 400 m. This means that the entire track is 400 m long.

### Step 2: Calculate the Time Taken for Each Distance
The athlete runs at a uniform speed of 15 m/s. We can calculate the time taken to run 100 m and 200 m.

- **For 100 m:**
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100 \text{ m}}{15 \text{ m/s}} = \frac{20}{3} \text{ seconds} \approx 6.67 \text{ seconds} \]

- **For 200 m:**
\[ \text{Time} = \frac{200 \text{ m}}{15 \text{ m/s}} = \frac{40}{3} \text{ seconds} \approx 13.33 \text{ seconds} \]

### Step 3: Determine the Change in Velocity for 100 m
1. **Initial Velocity (V1):** The initial velocity vector (V1) is directed tangentially at the starting point (let’s say point A).
2. **Final Position after 100 m:** The athlete moves 100 m along the circular track. Since the circumference is 400 m, moving 100 m means the athlete is at a quarter of the circle (point B).
3. **Final Velocity (V2):** The final velocity vector (V2) is also directed tangentially at point B.
4. **Direction of V1 and V2:** V1 is directed upwards (let's assume) and V2 is directed to the left at point B.

5. **Change in Velocity (ΔV):**
\[ \Delta V = V2 - V1 \]
Since V1 and V2 are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the change in velocity:
\[ |\Delta V| = \sqrt{V1^2 + V2^2} = \sqrt{(15 \text{ m/s})^2 + (15 \text{ m/s})^2} = \sqrt{225 + 225} = \sqrt{450} \approx 21.21 \text{ m/s} \]

### Step 4: Determine the Change in Velocity for 200 m
1. **Initial Velocity (V1):** The initial velocity vector (V1) is still directed tangentially at point A.
2. **Final Position after 200 m:** The athlete moves 200 m along the circular track, which means he is at the opposite side of the circle (point C).
3. **Final Velocity (V2):** The final velocity vector (V2) is directed tangentially at point C, which is in the opposite direction to V1.

4. **Change in Velocity (ΔV):**
\[ \Delta V = V2 - V1 \]
Since V1 and V2 are in opposite directions, we can simply add their magnitudes:
\[ |\Delta V| = |V2| + |V1| = 15 \text{ m/s} + 15 \text{ m/s} = 30 \text{ m/s} \]

### Final Answers:
- Change in velocity after running 100 m: **21.21 m/s**
- Change in velocity after running 200 m: **30 m/s**