Question

Two fishing boats leave a marina at the same time. One boat travels north at a speed of 12 miles per hour. The other boat heads east at a speed of 9 miles per hour. Approximately how far apart are the two boats 15 minutes after they leave the marina?

Answer 1

Answer:

5.25 miles

Step-by-step explanation:

15 min=1/4 of hour

12/4=3

9/4=2.25

3+2.25=5.25

Answer 2

Answer:

d = 4.59 miles

Step-by-step explanation:

Distance Between Two Points in the Plane

Given two points A(x,y) and B(w,z), the distance between them is:

$d=\sqrt{(z-y)^2+(w-x)^2}$

The fishing boats leave the marina in perpendicular directions. The first boat travels north at v=12 mph for t=15 minutes. Converting to hours t=15/60=0.25 hours.

The distance covered by this boat is

d1 = 12*0.25 = 4 miles

The boat ends up at the point (0,4).

The second boat travels east at v=9  mph for the same time t=0.25 hours. The distance is:

d2= 9*0.25 = 2.25 miles

This boat ends up at the point (2.25,0)

The distance between them is

$d=\sqrt{(4-0)^2+(0-2.25)^2}=\sqrt{16+5.0625}=\sqrt{21.0625}$

Calculating the square root:

d = 4.59 miles