Rader Detection: Finding Time

[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:
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

Is this right? 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?
Please help
Thanks!

 

[srmichael]

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 ====> Should be y = -3(7.2) + 12 = -9.6
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 ) ====> Wrong slope here, too.

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

Is this right? 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?
Please help
Thanks!

Make the corrections noted above and try again.

 

[Chaim]

Make the corrections noted above and try again.

Ah, I see, ok thanks
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

 

[pappus]

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.

...

I assume that the row boat is actually a transformer which becomes suddenly a tug boat?

Ah, I see, ok thanks
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

1. You certainely made a sketch. (see attachment)

2. You are interested in the total distance of |ET| + |TL|. As you've calculated these points have the coordinates:
E(7.6, -9.6), T(4, 0), L(4, 8√(2)).

Now use the distance formula to evaluate |ET| and |TL|. I've got \(\displaystyle d = \frac{16}5 \cdot \sqrt{10} + 8 \sqrt{2} \approx 21.433\)

3. Using the definition of \(\displaystyle speed = \frac{distance}{time}\) I've got

\(\displaystyle time = \frac{21.433}{15}=1.429\ h = \boxed{1\ h \ 25\ min\ 44 \ sec}\)

 

[Chaim]

I assume that the row boat is actually a transformer which becomes suddenly a tug boat?



1. You certainely made a sketch. (see attachment)

2. You are interested in the total distance of |ET| + |TL|. As you've calculated these points have the coordinates:
E(7.6, -9.6), T(4, 0), L(4, 8√(2)).

Now use the distance formula to evaluate |ET| and |TL|. I've got \(\displaystyle d = \frac{16}5 \cdot \sqrt{10} + 8 \sqrt{2} \approx 21.433\)

3. Using the definition of \(\displaystyle speed = \frac{distance}{time}\) I've got

\(\displaystyle time = \frac{21.433}{15}=1.429\ h = \boxed{1\ h \ 25\ min\ 44 \ sec}\)

Ah!! I see now!
Yeah, my mistake, I accidently typed another different kind of boat, they're were all suppose to be rowboat XD
Thanks!
I was doing it from one straight line, but forgot that it turns!
Thank you very much!