Question

A boat leaves a dock at 3:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 4:00 PM. How many minutes after 3:00 PM were the two boats closest together? (Round your answer to the nearest minute.)

Answer 1

Answer:

approximately ≅ 22minutes that is 3:22:00pm

Step-by-step explanation:

The two boats are at the dock at 3pm and 4pm respectively; let’s write everything in  terms of a common start time. Let 3pm be the time t = 0. At time t = 0, the first boat  is at the dock and the second boat is at 15 km west of the dock.

At time t, the first boat has moved distance 20t (recall that distance is speed × time),  so it is distance 20t from the dock. The second boat has moved distance 15t toward the  dock, but it was originally distance 15 away from the dock, so it is now 15 − 15t away  from the dock.

At time t, the distance between the two boats is then (by the Pythagorean Theorem)

$d =\sqrt{(15- 15t)^{2} + (20t)^{2} }$

We want to minimize this quantity.we therefore need to minimize;

$d^{2} = f(t) = (15 - 15t)^{2} + (20t)^{2} = 625t^{2} - 450t + 225.$

To minimize, we note that f ¹ (t) = 1250t − 450, so f ¹(t) = 0 occurs only at the time  t = $\frac{450}{1250}$ = $\frac{9}{25}$ = 0.36. This is the critical number, and since

f ¹¹(t) = 1250 > 0, it is indeed  a minimum.

So the boats are closest together at time t = 0.36, which is 0.36 hours (or 21.6 minutes)  approximately  22minutes

after 3pm.

that is 3:22:00pm

Diagram attached to the question for interpretation

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Answer 2

Answer:

The distance is minimized 22 minutes after 3:00 pm.

Step-by-step explanation:

Let t is the time for the 2 boats after 3:00 pm, we have the distance of:

• Boat 1: 20t
• Boat 2: 15t and its distance to the dock is given by (15-15t)

The paths of the boats form a right triangle, in which the hypotenuse (also the distance between the two boats) is:

D = (20t)² + (15 - 15t)²

<=> dD/dt = -2(15² )( 1-t ) +2 × 20² × t

<=> dD/dt = 2 (15² + 20²) × t -2 ( 15 )² = 0

<=> t = 2 ( 15 )² : 2 (15² + 20²)  

<=> t = 0.36 hours = 0.36 × 60 = 21.6 minutes ≈ 22 minutes

Therefore, the distance is minimized 22 minutes after 3:00 pm.