To solve this problem, we need to determine how far the faster car must travel before it has a 15-minute (or 1/4 hour) lead on the slower car.

Step-by-step solution:

1. Define the variables:

   • Speed of the slower car: $v_1 = 55 \, \text{mi/h}$
   • Speed of the faster car: $v_2 = 70 \, \text{mi/h}$
   • Time difference between the two cars: $t = 15 \, \text{minutes} = \frac{1}{4} \, \text{hour}$

2. Relative speed: The faster car gains on the slower car at a relative speed of: $v_{\text{relative}} = v_2 - v_1 = 70 - 55 = 15 \, \text{mi/h}$

3. Distance lead: The distance the faster car must travel to have a 15-minute (or $\frac{1}{4}$ hour) lead is the distance the slower car travels in that time, which is: $\text{Distance} = v_{\text{relative}} \times t = 15 \times \frac{1}{4} = 3.75 \, \text{miles}$

Conclusion:

The faster car must travel 3.75 miles before it has a 15-minute lead on the slower car.

Do you want more details, or have any questions?

Here are 5 related questions:

1. What if the faster car was traveling at 80 mi/h?
2. How long would it take for the faster car to gain a 30-minute lead?
3. If both cars start at the same time, how long will it take the faster car to gain a 5-mile lead?
4. What if the slower car's speed was 60 mi/h instead of 55 mi/h?
5. How far would both cars travel in 2 hours?

Tip: Converting time to hours when working with speeds in miles per hour ensures consistency in your calculations!