I'm having a little trouble with this problem...

Please try and walk me through the problem, too:

#1. A motorboat leaves a dock and travels at an average speed of 15 mi/h toward an island. The average speed on the return trip is 10 mi/h. How far is the island from the harbor if the total trip takes five hours?

#2. A car traveling at 88 km/h overtakes a bus traveling at 64 km/h. If the bus has a 1.5 hour head start, how far from the starting point does the car overtake the bus?

Thank you so much! Any help is appreciated!

Sure! Let's walk through each problem together and find the answers step by step.

#1. A motorboat leaves a dock and travels at an average speed of 15 mi/h toward an island. The average speed on the return trip is 10 mi/h. How far is the island from the harbor if the total trip takes five hours?

To find the distance between the harbor and the island, we need to use the formula: Distance = Speed × Time.

Let's start by assigning variables:
Let D be the distance between the harbor and the island.
Let t1 be the time taken to reach the island.
Let t2 be the time taken to return to the harbor.

We're given that the average speed towards the island is 15 mi/h, so we have:
t1 = D / 15 (from the equation Distance = Speed × Time)

Similarly, the average speed on the return trip is 10 mi/h, so we have:
t2 = D / 10

Now we know that the total trip takes 5 hours. So the sum of the time taken to reach the island and the time taken to return to the harbor should be 5 hours:
t1 + t2 = 5

We can substitute the values of t1 and t2 from the previous equations:
D / 15 + D / 10 = 5

To solve for D, we'll need to find a common denominator for both fractions. In this case, the common denominator is 30:
(2D + 3D) / 30 = 5

Adding the fractions together, we get:
5D / 30 = 5

Now we can solve for D by multiplying both sides of the equation by 30 and then dividing by 5:
5D = 150
D = 30

So, the distance between the harbor and the island is 30 miles.

#2. A car traveling at 88 km/h overtakes a bus traveling at 64 km/h. If the bus has a 1.5-hour head start, how far from the starting point does the car overtake the bus?

To find the distance from the starting point where the car overtakes the bus, we need to calculate the time it takes for the car to catch up to the bus.

Let's assign variables:
Let t be the time taken for the car to catch up to the bus.
Let d be the distance from the starting point where the car overtakes the bus.

Since the bus started 1.5 hours before the car, the bus has a head start. So, in 1.5 hours, the bus has traveled a distance of 1.5 hours multiplied by its speed:
Distance traveled by the bus = 1.5 × 64 = 96 km

Since the car is traveling faster than the bus, it will eventually catch up. When it does, the distance traveled by the car will be equal to the distance traveled by the bus plus the distance from the starting point where the car overtakes the bus:
Distance traveled by the car = Distance traveled by the bus + Distance from starting point

Now we can set up an equation using the formula Distance = Speed × Time:
88t = 96 + d

To solve for d, we need to find the value of t when the car catches up to the bus. Since both vehicles are traveling at different speeds, the ratio of their speeds will be equal to the ratio of the distances traveled:
64t / 64 = (t + 1.5) / 88

Simplifying the equation:
64t / 64 = (t + 1.5) / 88

Now, we can solve for t by cross-multiplying and simplifying the equation:
88 * 64t = 64 * (t + 1.5)

5632t = 64t + 96

Simplifying further:
5632t - 64t = 96

5568t = 96

Dividing both sides by 5568:
t = 96 / 5568

Now that we have t, we can substitute it back into the equation to find d:
d = 88t - 96

Calculating d:
d = 88 * (96 / 5568) - 96

Simplifying:
d = 1536 / 193 - 96

d = 7.961 km

Therefore, the car overtakes the bus approximately 7.961 km from the starting point.

I hope this explanation helps! Let me know if you have any further questions.