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1 year ago

8

Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes

and Boat 1 now sails at 9 knots. At 3 pm, the wind increases for Boat 2 and it now sails 7 knots faster. How fast (in knots) is the distance between the two ships changing at 5 pm. (Note: 1 knot is a speed of 1 nautical mile per hour.)

1 answer:

Ivan1 year ago

6 0

Answer:

14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

Now, speed of boat 1 changes to 8+7 = 15 knots

At 5 pm:

Distance traveled by boat 1 = 33 + 2 $\times$ 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 $\times$ 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled $\triangle AOB$ .

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units$

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = $\dfrac{Total\ distance}{Total\ time}$

$\Rightarrow \dfrac{74.28}{5} = 14.86\ knots$

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