Calculating Average Speed for a Round Trip Journey

The question asks us to find the average speed of a train during a complete round trip journey between two stations, A and B.

Average speed is defined as the total distance covered divided by the total time taken for the journey. It is important to distinguish average speed from average velocity, especially in a round trip where the displacement is zero.

Understanding the Journey Details

• Distance between Station A and Station B: 700 km
• Speed from A to B: 80 km/h
• Speed from B to A: 65 km/h

Step-by-Step Calculation of Average Speed

1. Calculate Total Distance

The train travels from A to B and then back from B to A. So, the total distance is the sum of the distance for the forward journey and the distance for the return journey.

Distance A to B = 700 km

Distance B to A = 700 km

Total Distance = Distance A to B + Distance B to A

Total Distance = 700 km + 700 km = 1400 km

We can represent this mathematically as:

\( \text{Total Distance} = D_{AB} + D_{BA} = 700 \, \text{km} + 700 \, \text{km} = 1400 \, \text{km} \)

2. Calculate Time Taken for Each Leg of the Journey

Time taken for a journey is calculated using the formula: Time = Distance / Speed.

Time taken from A to B (T_AB):

Distance A to B = 700 km

Speed A to B = 80 km/h

\( T_{AB} = \frac{\text{Distance A to B}}{\text{Speed A to B}} = \frac{700 \, \text{km}}{80 \, \text{km/h}} \)

\( T_{AB} = \frac{70}{8} \, \text{hours} = \frac{35}{4} \, \text{hours} = 8.75 \, \text{hours} \)

Time taken from B to A (T_BA):

Distance B to A = 700 km

Speed B to A = 65 km/h

\( T_{BA} = \frac{\text{Distance B to A}}{\text{Speed B to A}} = \frac{700 \, \text{km}}{65 \, \text{km/h}} \)

\( T_{BA} = \frac{140}{13} \, \text{hours} \)

3. Calculate Total Time Taken for the Whole Journey

Total Time = Time A to B + Time B to A

\( \text{Total Time} = T_{AB} + T_{BA} = \frac{35}{4} \, \text{hours} + \frac{140}{13} \, \text{hours} \)

To add these fractions, we find a common denominator, which is \(4 \times 13 = 52\).

\( \text{Total Time} = \frac{35 \times 13}{4 \times 13} + \frac{140 \times 4}{13 \times 4} = \frac{455}{52} + \frac{560}{52} \)

\( \text{Total Time} = \frac{455 + 560}{52} = \frac{1015}{52} \, \text{hours} \)

4. Calculate Average Speed

Average Speed = Total Distance / Total Time

\( \text{Average Speed} = \frac{1400 \, \text{km}}{\frac{1015}{52} \, \text{hours}} \)

\( \text{Average Speed} = 1400 \times \frac{52}{1015} \, \text{km/h} \)

Now, we perform the multiplication and division:

\( 1400 \times 52 = 72800 \)

\( \text{Average Speed} = \frac{72800}{1015} \, \text{km/h} \)

Dividing 72800 by 1015:

\( 72800 \div 1015 \approx 71.7241379... \)

The average speed is approximately 71.72 km/h.

Comparing with Options

Let's compare our calculated average speed (approximately 71.72 km/h) with the given options:

• 74 km/h
• 71.72 km/h
• 72.7 km/h
• 70.73 km/h

Our calculated value is very close to 71.72 km/h.

Conclusion on Average Speed

The average speed of the train during the whole journey is found by dividing the total distance (1400 km) by the total time taken (\(\frac{1015}{52}\) hours). The calculated value is approximately 71.72 km/h, which matches one of the provided options.

Revision Table: Key Concepts for Average Speed

Additional Information: Average Speed in Different Scenarios

It's useful to understand how average speed differs in various situations.

• Constant Speed: If an object moves at a constant speed, its average speed over any time interval is equal to that constant speed.
• Varying Speed: When the speed changes, we must calculate the total distance and total time. We cannot simply average the speeds (like \((80+65)/2\)) unless the time intervals for each speed are equal. In this problem, the time intervals are different because the distances are the same but speeds are different.
• Harmonic Mean: For a journey of equal distance covered at two different speeds, \(v_1\) and \(v_2\), the average speed is given by the harmonic mean of the speeds: \( v_{avg} = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}} = \frac{2v_1v_2}{v_1+v_2} \). Let's check this formula with our values: \( v_{avg} = \frac{2 \times 80 \times 65}{80 + 65} = \frac{2 \times 80 \times 65}{145} = \frac{10400}{145} \). \( \frac{10400}{145} = \frac{2080}{29} \approx 71.724 \), which matches our detailed calculation. This formula is a shortcut for equal distances.