Two boats traveling the same direction leave a harbor at noon. After 3 hr they are 60 miles apart, if one boat travels twice as fast as the other find the average rate of each boat

The boat 1 moves with a speed of 40mi/h, and boat 2 moves with a speed of 20mi/h.

Step-by-step explanation:

First, we know the relation:

Distance = Speed*Time.

We can define the average rate of the boats as the average speed of the boats.

Now, we know that two boats travel in the same direction, let's define:

S₁ = speed of boat 1.

S₂ = speed of boat 2.

We know that one travels twice as fast as the other, then we can write:

S₁ = 2*S₂

We also know that after 3 hours of travel, they are 60mi apart, then if the slower one travelled a distance D in 3 hours, then:

S₂*3h = D

And the faster one will travel D + 60mi

S₁*3h = (D + 60mi)

Then we have the equations:

S₂*3h = D

S₁*3h = (D + 60mi)

We can replace S₁ by 2*S₂ to get:

S₂*3h = D

(2*S₂)*3h = (D + 60mi)

Now we have isolated D in the above equation, we can just replace it in the second equation to get:

(2*S₂)*3h = (S₂*3h + 60mi)

Now we can solve this for S₂

S₂*6h = S₂*3h + 60mi

S₂*6h - S₂*3h = 60mi

S₂*3h = 60mi

S₂ = 60mi/3h = 20mi/h

The speed of boat 2 is 20mi/h

And we knew that:

S₁ = 2*S₂

then:

S₁ = 2*(20mi/h) = 40mi/h

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step-by-step explanation:

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