A car moves a distance of 100 m. It coveres first half with speed of 10 m/s and another half with speed of v m/s. If the average speed of car is 18 m/s, then value of v is

Options

A. 60 m/s
B. 80 m/s
C. 90 m/s
D. 110 m/s

Answer

To find the value of \( v \) in the given problem, we can follow these steps:

### Step 1: Understand the problem
The car travels a total distance of 100 meters. It covers the first half (50 meters) at a speed of 10 m/s and the second half (50 meters) at an unknown speed \( v \) m/s. We need to find the value of \( v \) given that the average speed for the entire journey is 18 m/s.

### Step 2: Use the formula for average speed
The average speed is defined as:
\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]
Given that the total distance is 100 meters and the average speed is 18 m/s, we can set up the equation:
\[ 18 = \frac{100}{T_{\text{total}}} \]

### Step 3: Calculate total time
From the equation above, we can find the total time \( T_{\text{total}} \):
\[ T_{\text{total}} = \frac{100}{18} \approx 5.56 \text{ seconds} \]

### Step 4: Calculate time for each half of the journey
1. **Time for the first half (50 meters at 10 m/s)**:
\[ T_1 = \frac{50}{10} = 5 \text{ seconds} \]

2. **Time for the second half (50 meters at \( v \) m/s)**:
\[ T_2 = \frac{50}{v} \]

### Step 5: Set up the equation for total time
The total time for the journey is the sum of the times for both halves:
\[ T_{\text{total}} = T_1 + T_2 \]
Substituting the values we have:
\[ \frac{100}{18} = 5 + \frac{50}{v} \]

### Step 6: Solve for \( v \)
Now we can solve the equation:
\[ \frac{100}{18} - 5 = \frac{50}{v} \]
Calculating \( \frac{100}{18} \):
\[ \frac{100}{18} \approx 5.56 \]
So,
\[ 5.56 - 5 = \frac{50}{v} \]
This simplifies to:
\[ 0.56 = \frac{50}{v} \]
Now, cross-multiplying gives:
\[ 0.56v = 50 \]
Solving for \( v \):
\[ v = \frac{50}{0.56} \approx 89.29 \text{ m/s} \]

### Step 7: Final answer
Thus, the speed \( v \) for the second half of the journey is approximately \( 89.29 \) m/s.