A train moving at a constnt velocity of 54 km/hr moves eastwards for 30 minuts, then due north with the same speed for 40 minutes. What is the average velocity of the train during this run? (in km/hr)

Options

A. 30
B. 35
C. `38.6`
D. `49.3`

Answer

To solve the problem of finding the average velocity of the train, we will follow these steps:

### Step 1: Convert the speed from km/hr to a more manageable unit for calculations.
The speed of the train is given as 54 km/hr.

### Step 2: Determine the distance traveled in the eastward direction.
The train moves east for 30 minutes. We convert 30 minutes into hours:
\[ 30 \text{ minutes} = \frac{30}{60} \text{ hours} = 0.5 \text{ hours} \]
Now, we can calculate the distance traveled east (D1):
\[ D1 = \text{Speed} \times \text{Time} = 54 \text{ km/hr} \times 0.5 \text{ hr} = 27 \text{ km} \]

### Step 3: Determine the distance traveled in the northward direction.
The train then moves north for 40 minutes. We convert 40 minutes into hours:
\[ 40 \text{ minutes} = \frac{40}{60} \text{ hours} = \frac{2}{3} \text{ hours} \approx 0.67 \text{ hours} \]
Now, we can calculate the distance traveled north (D2):
\[ D2 = \text{Speed} \times \text{Time} = 54 \text{ km/hr} \times \frac{2}{3} \text{ hr} = 36 \text{ km} \]

### Step 4: Calculate the total displacement.
The total displacement can be found using the Pythagorean theorem since the train travels in two perpendicular directions (east and north):
\[ \text{Displacement} = \sqrt{D1^2 + D2^2} = \sqrt{27^2 + 36^2} \]
Calculating the squares:
\[ 27^2 = 729, \quad 36^2 = 1296 \]
Adding them together:
\[ \text{Displacement} = \sqrt{729 + 1296} = \sqrt{2025} = 45 \text{ km} \]

### Step 5: Calculate the total time taken.
The total time taken for the journey is the sum of the time spent traveling east and north:
\[ \text{Total Time} = 0.5 \text{ hr} + \frac{2}{3} \text{ hr} = \frac{3}{6} + \frac{4}{6} = \frac{7}{6} \text{ hr} \approx 1.17 \text{ hr} \]

### Step 6: Calculate the average velocity.
Average velocity is defined as total displacement divided by total time:
\[ \text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{45 \text{ km}}{\frac{7}{6} \text{ hr}} = 45 \times \frac{6}{7} = \frac{270}{7} \approx 38.57 \text{ km/hr} \]

### Final Answer:
The average velocity of the train during this run is approximately **38.57 km/hr**.

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