Average Speed Calculation for Multiple Trips

This problem asks us to find the average speed of Dinesh over four trips of equal distance, where his speed changes for each trip.

Understanding Average Speed

The average speed is not simply the average of the speeds on each trip. It is calculated as the total distance traveled divided by the total time taken.

Formula for Average Speed:

\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]

Calculating Speed for Each Trip

We are given that the speed on the first trip is 140 km/hr, and the speed on each subsequent trip is half of the previous trip's speed.

• Speed on Trip 1 = 140 km/hr
• Speed on Trip 2 = \( \frac{140}{2} = 70 \) km/hr
• Speed on Trip 3 = \( \frac{70}{2} = 35 \) km/hr
• Speed on Trip 4 = \( \frac{35}{2} = 17.5 \) km/hr

Calculating Total Distance and Total Time

Let's assume the distance of each trip is \(d\) kilometers. Since there are four trips of equal distance, the total distance is:

\[ \text{Total Distance} = d + d + d + d = 4d \text{ km} \]

Now, we need to find the time taken for each trip. The relationship between distance, speed, and time is:

\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]

• Time for Trip 1 (\(t_1\)) = \( \frac{d}{140} \) hours
• Time for Trip 2 (\(t_2\)) = \( \frac{d}{70} \) hours
• Time for Trip 3 (\(t_3\)) = \( \frac{d}{35} \) hours
• Time for Trip 4 (\(t_4\)) = \( \frac{d}{17.5} = \frac{d}{(35/2)} = \frac{2d}{35} \) hours

The total time taken for all four trips is the sum of the time taken for each trip:

\[ \text{Total Time} = t_1 + t_2 + t_3 + t_4 = \frac{d}{140} + \frac{d}{70} + \frac{d}{35} + \frac{2d}{35} \]

To add these fractions, we find a common denominator, which is 140.

\[ \text{Total Time} = \frac{d}{140} + \frac{d \times 2}{70 \times 2} + \frac{d \times 4}{35 \times 4} + \frac{2d \times 4}{35 \times 4} \]

\[ \text{Total Time} = \frac{d}{140} + \frac{2d}{140} + \frac{4d}{140} + \frac{8d}{140} \]

\[ \text{Total Time} = \frac{d + 2d + 4d + 8d}{140} = \frac{15d}{140} \]

We can simplify the fraction \( \frac{15}{140} \) by dividing both numerator and denominator by 5.

\[ \text{Total Time} = \frac{15 \div 5}{140 \div 5} d = \frac{3d}{28} \text{ hours} \]

Calculating Average Speed

Now we can calculate the average speed using the total distance and total time:

\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{4d}{\frac{3d}{28}} \]

To divide by a fraction, we multiply by its reciprocal:

\[ \text{Average Speed} = 4d \times \frac{28}{3d} \]

The variable \(d\) cancels out:

\[ \text{Average Speed} = 4 \times \frac{28}{3} = \frac{4 \times 28}{3} = \frac{112}{3} \text{ km/hr} \]

Final Answer

The average speed of Dinesh in these four trips is \( \frac{112}{3} \) km/hr.

Revision Table: Key Concepts

Additional Information: Harmonic Mean

When a person travels equal distances at different speeds, the average speed is the harmonic mean of the speeds. For 'n' equal distance trips with speeds \(v_1, v_2, ..., v_n\), the average speed \(V_{avg}\) is given by:

\[ \frac{n}{V_{avg}} = \frac{1}{v_1} + \frac{1}{v_2} + ... + \frac{1}{v_n} \]

In this case, n = 4, and the speeds are 140, 70, 35, and 17.5 (or 35/2).

\[ \frac{4}{V_{avg}} = \frac{1}{140} + \frac{1}{70} + \frac{1}{35} + \frac{1}{17.5} \]

\[ \frac{4}{V_{avg}} = \frac{1}{140} + \frac{2}{140} + \frac{4}{140} + \frac{1}{35/2} = \frac{1}{140} + \frac{2}{140} + \frac{4}{140} + \frac{2}{35} \]

\[ \frac{4}{V_{avg}} = \frac{1}{140} + \frac{2}{140} + \frac{4}{140} + \frac{8}{140} = \frac{1+2+4+8}{140} = \frac{15}{140} = \frac{3}{28} \]

\[ \frac{4}{V_{avg}} = \frac{3}{28} \]

\[ V_{avg} = \frac{4 \times 28}{3} = \frac{112}{3} \text{ km/hr} \]

This confirms the result obtained using the total distance/total time method. The harmonic mean is a useful shortcut when dealing with average speed over equal distances.

It is important to remember that the simple arithmetic mean of the speeds (\((140+70+35+17.5)/4\)) would give an incorrect average speed in this scenario because the time spent at each speed is different.