During the first `18 min` of a `60 min` trip, a car has an average speed of `11 ms^-1.` What should be the average speed for remaining `42 min` so that car is having an average speed of `21 ms^-1` for the entire trip?

Options

A. `25.3 ms^-1`
B. `29.2 ms^-1`
C. `31 ms^-1`
D. `35.6 ms^-1`

Explanation

To solve the problem step by step, we need to find the average speed required for the remaining 42 minutes of the trip so that the overall average speed for the entire 60-minute trip is 21 m/s.

### Step 1: Calculate the distance covered in the first 18 minutes.
- The average speed during the first 18 minutes is given as 11 m/s.
- First, convert 18 minutes into seconds:
\[ 18 \text{ minutes} = 18 \times 60 = 1080 \text{ seconds} \]
- Now, calculate the distance \(d_1\) covered in the first 18 minutes:
\[ d_1 = \text{speed} \times \text{time} = 11 \, \text{m/s} \times 1080 \, \text{s} = 11880 \, \text{meters} \]

### Step 2: Set up the equation for the total distance and average speed.
- The total time for the trip is 60 minutes, which is:
\[ 60 \text{ minutes} = 60 \times 60 = 3600 \text{ seconds} \]
- The desired average speed for the entire trip is 21 m/s.
- The total distance \(D\) for the entire trip can be calculated using the average speed:
\[ D = \text{average speed} \times \text{total time} = 21 \, \text{m/s} \times 3600 \, \text{s} = 75600 \, \text{meters} \]

### Step 3: Calculate the distance for the remaining 42 minutes.
- Let \(d_2\) be the distance covered in the remaining 42 minutes.
- The distance \(d_2\) can also be expressed in terms of the average speed \(V\) for the remaining time:
\[ d_2 = V \times \text{time} = V \times (42 \times 60) = V \times 2520 \, \text{seconds} \]

### Step 4: Set up the equation for total distance.
- The total distance \(D\) is the sum of the distances covered in both segments:
\[ D = d_1 + d_2 \]
- Substituting the known values:
\[ 75600 = 11880 + 2520V \]

### Step 5: Solve for \(V\).
- Rearranging the equation gives:
\[ 2520V = 75600 - 11880 \]
- Calculate the right-hand side:
\[ 2520V = 63420 \]
- Now, divide both sides by 2520 to find \(V\):
\[ V = \frac{63420}{2520} \approx 25.2 \, \text{m/s} \]

### Conclusion
- The average speed required for the remaining 42 minutes is approximately \(25.2 \, \text{m/s}\).