A submarine kept a constant rate of ascent upon rising after a dive. After 5 minutes it was 360 feet below the surface of the water, and after 25 minutes it was 140 feet below. How long after starting its ascent was the submarine 250 feet below the surface?

In 20 minutes it rose 220ft

So, the depth is decreasing at a rate of 11 ft/min

360 = 415 - 11*5, so the depth

d(t) = 415 - 11t

So, when d=250,

415-11t = 250
11t = 165
t = 15

So, 15 minutes after starting, the depth was 250 ft.

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To solve this problem, we can use the concept of proportionality between time and depth.

Let's first define the time taken for the submarine to reach the surface as "t" (in minutes), and the depth (in feet) at that time as "d."

From the given information, we know that after 5 minutes, the submarine is 360 feet below the surface, and after 25 minutes, it is 140 feet below the surface.

Now, we can set up a proportion between the time and depth:

(25 - 5)/(140 - d) = 25/t

Simplifying this equation, we get:

20/(140 - d) = 1/t

Cross-multiplying, we have:

20t = 140 - d

Simplifying further:

d = 140 - 20t

We want to find the time (t) when the submarine is at a depth of 250 feet below the surface. Substituting d = 250 into the equation, we have:

250 = 140 - 20t

Adding 20t to both sides:

20t = 140 - 250

20t = -110

Dividing both sides by 20:

t = -110/20

t = -5.5

This implies that after 5.5 minutes of starting its ascent, the submarine would be 250 feet below the surface.

Since time cannot be negative, we would disregard the negative value and conclude that the submarine would be 250 feet below the surface after approximately 5.5 minutes of starting its ascent.