Verified answer

Yes, the poor Creature can survive running directly North to the point (99, 10)! The rules require the Villain to keep direction to the Victim and that will save it, indeed for time "t" Victim would be 2t North of the starting point and if y = y(x) is the equation of the Villain's persecution curve, the distance covered by him would be

20t = ∫[0, x] sqrt(1 + y'²) dx, hence

dt/dx = sqrt(1+y'²)/20, on the other hand the Villain's velocity vector has the direction of the tangent to y = y(x), this tangent intersecting x-axis at x - y/y'. Taking tangent from this point, x-axis and the vertical lines ξ=x and ξ=99 we have 2 similar right triangles, so:

2t/y = (99 - (x - y/y'))/(x - (x - y/y')), or

2t = y + (99 - x)y', differentiating that

dt/dx = y " (99 - x)/2, so for the persecution curve y = y(x) we have the 2nd order differential equation:

y " (99 - x)/2 = sqrt(1 + y'²)/20, or

y " (99 - x) = sqrt(1 + y'²)/10, initial conditions:

y(0) = 0, y'(0) = 0 /starting point - origin, starting direction - East/. Solving it by substitution z = y', we obtain:

∫dz/sqrt(1 + z²)) = (1/10)∫dx/(99 - x) + const, or

(z + sqrt(1 + z²))*(1 - x/99)^(1/10) = 1, back to z = y' yields

2y' = (1 - x/99)^(-1/10) - (1 - x/99)^(1/10) and finally

y = 45(1 - x/99)^(11/10) - 55(1 - x/99)^(9/10) + 10

This is the equation of the Villain's path. Taking x=99 here we obtain y(99) = 10, so both the Villain and Victim will meet in the point (99, 10) after exactly 99*20/(20² - 2²) = 5 seconds, but, according the rules of the persecution, now

y = 10, so the Villain can do nothing more than to scream helplessly.

Cheers for the Victims!

P.S.(EDIT - after having read previous answers) Both David and Emmet are right about the chicken's optimal strategy - running straight North, but the explanations are not satisfactory: the persecution path is a curve indeed, so at the 1st second Reaper's position is NOT (20, 0). Neither at the 5th second it will be (99.83, 5.03), here the Emmet's calculations /I can't read them, Yahoo Answers has cut them/ are wrong, the final position is exactly (99, 10) and the velocities ratio 20:2 is critical in this problem - the chicken WILL NOT BE SAFE if its speed is less.

Generally, all the data in this really nice and interesting problem are excellently chosen to match the optimal strategy.

the chicken could just go straight up without getting run over. I think, i considered it by counting the distances they covered per second. If so, the chicken just going straight up would be safe within 5 seconds.

Taking note of 5 seconds, I considered the reaper, from (0,0), he'd go to (20,0) in the first second since his target, the chicken is at y=0.

By the 2nd second (kinda awkward grammar), the case would already change. I assumed a right triangle, assuming that the reaper at (20,0) would already be facing the chicken at y=2. so i got 79 for its length (chicken at 29 minus reaper at 20) and height at 2.

I used the tangent function to solve for the angle adjacent the 79 units long side of the triangle, tan(angle) = 2(opposite side of the reference angle)/9(adjacent side of the reference angle).

I got an angle of about 1.45

Using this angle, and a hypotenuse of 20, I got an x value of about 19.99 and a y value of 0.5.

for every second until the 5th second, the reaper would need to change his angle with respect to the chicken's postion.

for the 2nd up to the 5th second i got these angles

2 sec - 1.45 degrees

3 sec - 1.94

4 sec - 2.93

5 sec - 12.52

now, if we compute for the distance covered by the reaper in terms of x, at 5 seconds, he would already be at about x =100. But do not fear, chicken!!! There is still the y component of the reaper...

for each second, the y incremental values for 1, 2, 3, 4 ,5 seconds are 0, .5, .67, 1.02, and 4.3

By the time the chicken is safe, at 5 sec and at (99,10), the reaper would only be at y = 6.49. the chicken is already way up.

I think it is safe to assume that second by second computation would still be accurate because in less than 1 second, the angle changes accordingly...

I think...

we can find the reapers velocity at any moment with the system of eqn's Vx=20Cos[tan-1(Py(c)/99)]

Vy=20Sin[tan-1(Py(c)/99)]

where Py(c) is the y position of the chicken

then from this we can find the eqn's for the position of the reaper by taking the integral of each side from 0 to the time when the chicken reaches y=10...i.e. 10/Vy(c)

Px(g)=int{20Cos[tan-1(Py(c)/99)], t, 0, 10/Py(c)}

Py(g)=int{20Sin[tan-1(Py(c)/99)], t, 0, 10/Py(c)}

solving this solution for differant initial velocities the chicken will definitely cross the road safely if it crosses in a straight line. the chicken will be at (99,10) while the reaper will be at (99.83,5.03). the chicken will be safe at any Vy(c) between 1.01 and 2 meters per second. further analysis may yield that with the reapers homing device and his tremendous speed he may not be able to catch the chicken on numerous courses due to overshooting.errors (notice in the straight line example the reaper already overshot the chicken in the x direction..;.)

Just a question to check whether I understood the tracking device.

If the chicken's path is directly north, then the reaper's path will be the hypotenuse of the right triangle formed by the chicken's path and the x-axis. Is that correct? Thanks.

Seeing as chickens can fly, and in general motorbikes can't, then I think the addition of a "z" axis will solve the chicken's dilemma...

stuff that! no humans i know have met the grimreaper, as if we care if a chicken has! Well just brush him off as a liar anyway.