Question

A horse rider covers half the distance with $5\, m / s$ speed The remaining part of the distance was travelled with speed $10\, m / s$ for half the time and with speed $15\, m / s$ for other half of the time The mean speed of the rider averaged over the whole time of motion is $\frac{x}{7} m / s$ The value of $x$ is _____

Answer 1

Correct Answer: 50

To find the mean speed $\frac{x}{7}\, m/s$ of the rider, we need to consider both the distance and time factors for each segment of the journey. Let's break down the problem:

1. **Distance Analysis**: Suppose the total distance is $D$.
- First half of the distance: \( \frac{D}{2} \), at $5\, m/s$.
- Remaining half: Also \( \frac{D}{2} \), but with different speed considerations.

2. **Time Calculation for Each Segment**:
- Time for first half: \( t_1 = \frac{\frac{D}{2}}{5} = \frac{D}{10} \).
- Let \( t_2 \) be the time taken to travel the second half: Half of this part is covered at $10\, m/s$, and the other half at $15\, m/s$:
  - Time for $10\, m/s$: \( t_{10} = \frac{t_2}{2} \)
  - Time for $15\, m/s$: \( t_{15} = \frac{t_2}{2} \)

3. **Distance Covered in Second Half**:
- Distance at $10\, m/s$: \( v_{10} \cdot \frac{t_2}{2} = 10 \cdot \frac{t_2}{2}\)
- Distance at $15\, m/s$: \( v_{15} \cdot \frac{t_2}{2} = 15 \cdot \frac{t_2}{2}\)
- Total distance for second half: \( \frac{D}{2} = \frac{10 \cdot t_2}{2} + \frac{15 \cdot t_2}{2}\).

4. **Solve for \( t_2 \):**
\[ \frac{D}{2} = 12.5 \cdot t_2 \implies t_2 = \frac{D}{25} \]

5. **Total Time \( T \):**
\[ T = t_1 + t_2 = \frac{D}{10} + \frac{D}{25} = \frac{5D + 2D}{50} = \frac{7D}{50} \]

6. **Mean Speed Calculation:**
- Mean speed is total distance divided by total time:
\[ v_{mean} = \frac{D}{T} = \frac{D}{\frac{7D}{50}} = \frac{50}{7}\, m/s \]
- Given mean speed: \( \frac{x}{7}\, m/s\)
- Equating the expressions: \(\frac{50}{7} = \frac{x}{7}\) implies \( x = 50 \).

The solution satisfies the range \(50,50\), confirming \( x \) is indeed 50.