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Answer:
<em>Roberto's speed in still water is 6 miles/hour and the river speed is 2 miles/hour</em>
Step-by-step explanation:
<u>Relative Speed</u>
When a body is moving at a constant speed v, the distance traveled in a time t is:
When Roberto rows downstream, his speed in still water is added to the speed of the water, making it easier to travel the required distance.
When Roberto rows upstream, his speed in still water is affected by the speed of the water, both are subtracted and the required distance is covered in more time.
Let's call
x = Roberto's rowing speed in still water
y = Speed of the river current
The speed when rowing downstream is x+y, thus the distance traveled is
Where t1=2.5 hours. Substituting values:
Rearranging, we find the downstream equation:
The speed when rowing upstream is x-y, and the distance traveled is
Where t2=5 hours. Substituting values:
Rearranging, we find the upstream equation:
Multiplying [1] by 2:
Adding this equation to [2]:
Solving:
Dividing [2] by 5:
Solving for y
Thus, Roberto's speed in still water is 6 miles/hour and the river speed is 2 miles/hour
Answer:
D = L/k
Step-by-step explanation:
Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is
dA/dt = in flow - out flow
Since litter falls at a constant rate of L grams per square meter per year, in flow = L
Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow
So,
dA/dt = in flow - out flow
dA/dt = L - Ak
Separating the variables, we have
dA/(L - Ak) = dt
Integrating, we have
∫-kdA/-k(L - Ak) = ∫dt
1/k∫-kdA/(L - Ak) = ∫dt
1/k㏑(L - Ak) = t + C
㏑(L - Ak) = kt + kC
㏑(L - Ak) = kt + C' (C' = kC)
taking exponents of both sides, we have
When t = 0, A(0) = 0 (since the forest floor is initially clear)
So, D = R - A =
when t = 0(at initial time), the initial value of D =