Understanding the Journey Problem

This problem involves calculating the total distance of a journey given the total time and the speeds for two equal halves of the distance. We need to use the fundamental relationship between distance, speed, and time.

The key information provided is:

• Total journey time: 10 hours
• Speed for the first half of the journey: 20 km/h
• Speed for the second half of the journey: 30 km/h

The first half and the second half of the journey cover the same distance.

Setting Up the Speed, Distance, Time Calculation

Let the total distance of the journey be \(D\) kilometres.

Since the journey is divided into two equal halves, the distance of the first half is \(\frac{D}{2}\) km, and the distance of the second half is also \(\frac{D}{2}\) km.

The relationship between speed, distance, and time is given by:

\(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\)

Calculating Time for Each Half

Let \(t_1\) be the time taken to complete the first half of the journey and \(t_2\) be the time taken to complete the second half of the journey.

For the first half:

• Distance = \(\frac{D}{2}\) km
• Speed = 20 km/h
• Time \(t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{D/2}{20} = \frac{D}{2 \times 20} = \frac{D}{40}\) hours

For the second half:

• Distance = \(\frac{D}{2}\) km
• Speed = 30 km/h
• Time \(t_2 = \frac{\text{Distance}}{\text{Speed}} = \frac{D/2}{30} = \frac{D}{2 \times 30} = \frac{D}{60}\) hours

Finding the Total Journey Distance

The total time for the journey is the sum of the time taken for the first half and the time taken for the second half. We are given that the total time is 10 hours.

\(\text{Total Time} = t_1 + t_2\)

\(10 = \frac{D}{40} + \frac{D}{60}\)

To solve for \(D\), we need to find a common denominator for 40 and 60. The least common multiple (LCM) of 40 and 60 is 120.

Multiply both sides of the equation by 120 to eliminate the denominators:

\(120 \times 10 = 120 \times \left(\frac{D}{40} + \frac{D}{60}\right)\)

\(1200 = 120 \times \frac{D}{40} + 120 \times \frac{D}{60}\)

\(1200 = 3D + 2D\)

\(1200 = 5D\)

Now, divide by 5 to find the value of \(D\):

\(D = \frac{1200}{5}\)

\(D = 240\)

So, the total journey he travelled is 240 kilometres.

Verification

Let's check if this distance results in a total time of 10 hours.

Distance of each half = \(\frac{240}{2} = 120\) km

Time for the first half (\(t_1\)) = \(\frac{120 \text{ km}}{20 \text{ km/h}} = 6\) hours

Time for the second half (\(t_2\)) = \(\frac{120 \text{ km}}{30 \text{ km/h}} = 4\) hours

Total time = \(t_1 + t_2 = 6 + 4 = 10\) hours.

This matches the given total time, so our calculation for the total distance is correct.

The total journey distance is 240 km.

Revision Table: Speed, Distance, Time Formulas

Additional Information: Average Speed for Journey Calculations

When dealing with journeys where the speed changes, the average speed is not simply the arithmetic mean of the speeds. If an object travels two equal distances \(d\) at speeds \(v_1\) and \(v_2\), the total distance is \(2d\) and the total time is \(\frac{d}{v_1} + \frac{d}{v_2}\). The average speed (\(v_{avg}\)) is:

\(v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2d}{\frac{d}{v_1} + \frac{d}{v_2}} = \frac{2d}{\frac{d(v_1 + v_2)}{v_1 v_2}} = \frac{2 v_1 v_2}{v_1 + v_2}\)

This is the harmonic mean of the two speeds. In our problem, \(v_1 = 20\) km/h and \(v_2 = 30\) km/h. The average speed would be:

\(v_{avg} = \frac{2 \times 20 \times 30}{20 + 30} = \frac{1200}{50} = 24\) km/h.

Using the average speed, the total distance \(D\) can also be found using \(D = v_{avg} \times \text{Total Time}\):

\(D = 24 \text{ km/h} \times 10 \text{ hours} = 240\) km.

This confirms our previous result and shows another way to approach problems involving average speed over equal distances.