1. ## Radar Detection: Finding time

A radar buoy detects any boats within a radius of 12 miles. A rowboat starts at a location 18 miles south and 10 miles east of the radar buoy. The rowboat travels at a constant speed of 15 mph. The tugboat travels on a straight line toward the northermost point of the radar region. When the tugboat is directly east of the buoy, it turns and travels due north until it exits the radar region.

How long (in hours) is the tugboat in the radar region?

This is what I did:
I represented it by a circle, letting the radar be the center (0,0)
Here are my steps:
1) x2 + y2 = 122
2) x2 + y2 = 144

The initial position P(10,-18) is heading to (0,12)

So I find the linear equation of the line:
3) Slope = (12-(-18))/(0-(10)) = -3 , with a
Y-intercept is 12
So first path: y = -3x + 12

Where it crosses
The circle:
5) x2 + (-3x+12)2 = 144
6) x2 + 9x2 + 72x + 144 = 144
7) 10x2+ 72x = 0
8) x(10x + 72) = 0
9) x = 0 or x = 7.2

So now I know x=0 is my point
Our point (0,12)
If x = 7.2
10) y = 3(7.2) + 12 = 33.6
So the intersection of his path with the radar region is (7.2, 33.6)
I crosses the x-axis at x = -4 , (let y=0 in y = 3x+12 )

So first leg is from (7.2 , 33.6) to (-4,0) , then it heads north
so let x = -4 in the circle equation:

16 + y2 = 144
y = √128 , which is the distance it travels northbound
within the region.

Finally find the distance between (-7.2, -9.6) and
(-4,0) , add on √128 for the total distance.
Since Time = distance/rate

Distance between (-7.2, -9.6) and (-4,0) would be
d=√(-4-(-7.2))2+(0-33.6)) = √((3.2)2+(-33.6)2)
d=1139.2
11) Time = 1139.2/15 = 75.94 hours

It seems WAY too much! (I tried getting help from some other people, and they say the answer is 5.5 hours in radar region, though I don't know how they got it so I don't know if they got it right).
Is my first leg correct?
Thanks!

2. ## Re: Radar Detection: Finding time

Originally Posted by Chaim
A radar buoy detects any boats within a radius of 12 miles. A rowboat starts at a location 18 miles south and 10 miles east of the radar buoy. The rowboat travels at a constant speed of 15 mph. The tugboat travels on a straight line toward the northermost point of the radar region. When the tugboat is directly east of the buoy, it turns and travels due north until it exits the radar region.

How long (in hours) is the tugboat in the radar region?

This is what I did:
....
So first path: y = -3x + 12 <--- OK

Where it crosses
...
9) x = 0 or x = 7.2 <--- OK

So now I know x=0 is my point
Our point (0,12)
If x = 7.2
10) y = 3(7.2) + 12 = 33.6 <--- this is wrong: The slope has the value -3
So the intersection of his path with the radar region is (7.2, 33.6)
I crosses the x-axis at x = -4 , (let y=0 in y = 3x+12 )

...
Draw a sketch! So you have an opportunity to control your arithmetic results.

3. ## Re: Radar Detection: Finding time

Ah ok, I saw my mistake
So when it's -3x+12
y=-3(7.2) + 12 = -9.6
Then I used the distance formula
d=sqrt((4+7.2)2)+(0+9.6)2))
d=sqrt(125.44+96.16)
d=sqrt(217.6)
d=14.7512711

Then for t
t=distance/rate
t=14.7512711/15
d=0.983418073 hours?

Did I do this right?
This seems way too less now