To solve this problem, let's denote the following:

• $S_1 = 40$ km/h (speed during the first part of the journey)
• $S_2$ = speed needed during the return journey
• $\text{Average Speed} = 60$ km/h
• $d$ = distance for each leg of the journey (same distance both ways)

The formula for average speed when the total distance is the same for both parts of the journey is:

$\text{Average Speed} = \frac{2 \times S_1 \times S_2}{S_1 + S_2}$

Given that the average speed should be 60 km/h, we can plug in the values:

$60 = \frac{2 \times 40 \times S_2}{40 + S_2}$

Now, solve for $S_2$:

1. Multiply both sides by $40 + S_2$:

$60 \times (40 + S_2) = 2 \times 40 \times S_2$

2. Distribute and simplify:

$2400 + 60S_2 = 80S_2$

3. Move all terms involving $S_2$ to one side:

$2400 = 80S_2 - 60S_2$

$2400 = 20S_2$

4. Solve for $S_2$:

$S_2 = \frac{2400}{20} = 120 \text{ km/h}$

So, the car needs to travel at 120 km/h on the return journey to achieve an average speed of 60 km/h for the entire trip.

Would you like more details or have any questions?

Here are five related questions:

1. What if the average speed required was 50 km/h instead?
2. How does the time taken for each leg of the journey relate to the speeds?
3. What happens to the average speed if the return journey speed is less than the first journey?
4. Can you calculate the total time taken for the journey with these speeds?
5. How would the problem change if the distances for the two parts of the journey were different?

Tip: Remember, when calculating average speed over two legs of a journey with different speeds, it’s not simply the arithmetic mean of the two speeds. Instead, it's based on the total distance and total time.