Line / Circle intersection

**actorRunning** · February 28th 2011, 11:55 AM

Alright, here's the problem:

Water is flowing from a major broken water main at the intersection of two streets. The resulting puddle of water is circular and the radius r of the puddle is given by the equation r = 5t feet, where t represents time in seconds elapsed since the main broke.

a) When the main broke, a runner was located 6 miles from the intersection. The runner continues toward the intersection at the constant speed of 17 feet per second. When will the runner's feet get wet?

b) Suppose, instead, that when the main broke, the runner was 6 miles east, and 5000 feet north of the intersection. The runner runs due west at 17 feet per second. When will the runner's feet get wet?

Here's what I've done:
Circle: $x^{2}+y^{2}=25t^{2}$
Runner: $Runner&:\;\;d_{from\;intersection}=6mi\cdot\frac{5 280ft}{mi}-\frac{17ft}{sec}\cdot t\;sec=31680-17t$
let y=0, then: $(31680-17t)^{2}=25t^{2}$
$t=1440\;sec=24\;minutes$

So part a) is simple. Part b), however, is giving me grief. Leaving the origin where it is, I'm able to see that the portion of the spill covering the street on which the runner is moving (5000 ft north) is encroaching on the runner according to $\sqrt{(5t)^{2}-5000^{2}}$ . It appears to me that I can equate that with the runner's motion towards the spill and solve for time, but when I do I end up with a NASTY quadratic that seems to fall apart when I try to solve it. The back of my textbook demurely states "Wet in 25.4154041 minutes," which sounds reasonable, but my path there right now is much less direct than the runner's.

Advise? Or direction pointing? Thank you so much!

**CaptainBlack** · February 28th 2011, 02:39 PM

Originally Posted by actorRunning

Alright, here's the problem:

Water is flowing from a major broken water main at the intersection of two streets. The resulting puddle of water is circular and the radius r of the puddle is given by the equation r = 5t feet, where t represents time in seconds elapsed since the main broke.

a) When the main broke, a runner was located 6 miles from the intersection. The runner continues toward the intersection at the constant speed of 17 feet per second. When will the runner's feet get wet?

b) Suppose, instead, that when the main broke, the runner was 6 miles east, and 5000 feet north of the intersection. The runner runs due west at 17 feet per second. When will the runner's feet get wet?

Here's what I've done:
Circle: $x^{2}+y^{2}=25t^{2}$
Runner: $Runner&:\;\;d_{from\;intersection}=6mi\cdot\frac{5 280ft}{mi}-\frac{17ft}{sec}\cdot t\;sec=31680-17t$
let y=0, then: $(31680-17t)^{2}=25t^{2}$
$t=1440\;sec=24\;minutes$

So part a) is simple. Part b), however, is giving me grief. Leaving the origin where it is, I'm able to see that the portion of the spill covering the street on which the runner is moving (5000 ft north) is encroaching on the runner according to $\sqrt{(5t)^{2}-5000^{2}}$ . It appears to me that I can equate that with the runner's motion towards the spill and solve for time, but when I do I end up with a NASTY quadratic that seems to fall apart when I try to solve it. The back of my textbook demurely states "Wet in 25.4154041 minutes," which sounds reasonable, but my path there right now is much less direct than the runner's.

Advise? Or direction pointing? Thank you so much!

For the second part you have for the runner:

$x_r=31680-17t$

and

$y_r=5000$

When this meets the expanding pool you have:

$x_r^2+y_r^2=20t^2$

So we are looking for the smallest root of:

$(31680-17t)^2+5000^2=25t^2$

which simplifies to:

$264\,{t}^{2}-1077120\,t+1028622400=0$

and the smallest root of this is $1524.92$ s or about $25.4$ min

CB

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