How did the chicken cross the road?

Make the following assumptions:

$d =$ the north/south distance between the next car and the chicken when the chicken starts crossing (in meters)

$t =$ the amount of time (in seconds) until the car reaches the chicken

$\theta =$ the angle the chicken crosses the road (radians north of east)

$w =$ the east/west distance the chicken travels before the car reaches it

$s_{car} =$ the speed of the car = 80 km/hr = 200/9 m/s

$s_{chicken} =$ the speed of the chicken = 0.25 m/s

Then we have the following:

$w = s_{chicken}t\cos\theta$

$s_{car}t = d + s_{chicken}t\sin\theta$

Solving for $t$ in the second equation we get:

$t = \frac {d} {s_{car} - s_{chicken}\sin\theta}$

Substituting this value of $t$ into the first equation we have:

$w = \frac {s_{chicken}d\cos\theta} {s_{car} - s_{chicken}\sin\theta}$

We want to maximize the chicken's chance of crossing the road successfully, which is equivalent to maximizing the value of $w$ . We do this by differentiating the RHS of the last equation with respect to $\theta$ and setting the derivative equal to 0.

$w' = \frac {-s_{chicken}d\sin\theta\times(s_{car} - s_{chicken}\sin\theta) - s_{chicken}d\cos\theta\times(- s_{chicken}\cos\theta)}{(s_{car} - s_{chicken}\sin\theta)^2}$

$= \frac {s_{chicken}^2d - s_{chicken}s_{car}d\sin\theta} {(s_{car} - s_{chicken}\sin\theta)^2}$

Setting this derivative equal to 0, we find that:

$\sin\theta = \frac {s_{chicken}} {s_{car}}$

$\sin\theta = \frac {0.25} {200/9}$

$\theta = 0.01125\space radians$

We can totally ignore how the cars are being distributed. This is because what we want is to maximize the chicken's chance of not getting hit by the car, we cant actually do anything about how the cars are distributed.

Here, i would like to define $\theta$ as the angle between the north-south axis and the velocity vector of the chicken.

Let the width of the car be $w$ , and imagine you're the driver of the car on this road. The time taken, t, for the chicken to reach the other side of the road assuming there are no cars present is

$t= \frac{w}{v \sin \theta}$

where v is the velocity of the chicken given in the question. To you, the driver, assuming you don't hit the chicken, you should be able to see that the distance (relative to you) that the chicken travels is

$s = \sqrt{ ( \frac{200}{9} - v \cos \theta )^{2} ( \frac{w}{v \sin \theta})^{2} + w^{2} }$
And that ^ is what we want to minimize.

Sigh. now we have to differentiate this huge stuff with respect to $\theta$ . But remember we just find out the solutions to $\frac{ds}{d \theta}$ =0.......

And not find out what $\frac{ds}{d \theta}$ actually is.

Alright, so let us take $(\frac{200}{9} - v \cos \theta )^{2} ( \frac{w}{v \sin \theta})^{2} + w^{2} = f( \theta)$

By chain rule you should see that if we wanna find the minimum of s, $\frac{ds}{d \theta} =\frac{f^{\prime}(\theta)}{ 2\sqrt{f(\theta)}} = 0$

Now we will solve that equation. $f^{\prime}(\theta) = 2w^{2} (\frac{ \frac{200}{9} - v\cos\theta}{vsin\theta}) \frac{v^{2}(sin \theta)^{2}+v^{2}(cos \theta)^{2}-\frac{200}{9}vcos\theta}{v^2(sin\theta)^{2}}$ =0

The solutions are now $\frac{200}{9} - vcos \theta = 0$ or $v=0$ or $v- \frac{200}{9}cos\theta =0$

Only the last solution makes sense if you look carefully. So this gives us easily $\theta = 1.559546........$

So to find the required answer, we just take $\frac{\pi}{2} - \theta = 0.01125 (4 sig fig)$

Suppose the width of the road is $w$ metres and the distance to the next car is $d$ metres. Furthermore, suppose that the chicken crosses at a straight line and makes an angle $\theta$ with respect to the horizontal (east). The time taken for the chicken to cross (should it still be alive) is $\frac{w}{0.25 \cos \theta}$ seconds. Noting that the speed of the cars in m/s is $\frac{200}{9}$ , then the distance the next car moved in the time it takes for the chicken to cross the road will be $\frac{200}{9} \times \frac{w}{0.25 \cos \theta}$ metres. Meanwhile, the chicken moves $w \tan \theta$ metres northwards as well. So the car would have closed up to the chicken by $\frac{200}{9} \times \frac{w}{0.25 \cos \theta} - w \tan \theta$ metres. We want this to be less than or equal to $d$ metres in order for the chicken not to be killed, so $\frac{200}{9} \times \frac{w}{0.25 \cos \theta} - w \tan \theta \leq d$ . This gives, after some simplification, $\frac{800}{9} \leq \frac{d}{w} \cos \theta + \sin \theta$ . Let $\frac{d}{w} = m$ . Then the right-hand side is equal to $\sqrt{m^2+1} \cos (\theta - \tan^{-1} \frac{1}{m})$ . So we get $\frac{800}{9\sqrt{m^2+1}} \leq \cos (\theta - \tan^{-1} \frac{1}{m})$ . Here, the right-hand side is not more than $1$ , so the chicken can survive only if $\frac{800}{9\sqrt{m^2+1}} \leq 1$ , i.e. $m \geq \sqrt{(\frac{800}{9})^2-1}$ . When $m = \sqrt{(\frac{800}{9})^2-1}$ , we have $\tan^{-1} \frac{1}{m} \approx 0.01125$ , so we need $\theta = 0.01125$ for the chicken to survive. Now, as we increase $m$ , then $\frac{800}{9\sqrt{m^2+1}}$ decreases much faster than $\cos (\theta - \tan^{-1} \frac{1}{m})$ , so the inequality will still hold for all $m > \sqrt{(\frac{800}{9})^2-1}$ . Therefore, when we set $\theta = 0.01125$ , the chicken will survive whenever it has a chance to (i.e. when $m \geq \sqrt{(\frac{800}{9})^2-1}$ ). So to maximize the chicken's chance of making it across the road, we let $\theta = 0.01125$ .

Let the width of the road be denoted by $w$ . Let the angle at which the chicken crosses be $\theta$ . The length $L$ of the path of the chicken across the road is then $w\sec\theta$ , and the distance $h$ in the North-South direction that the chicken travels is $w\tan\theta$ . Let $v_1=0.25 \, \mathrm{\frac{m}{s}}$ and $v_2=80 \, \mathrm{\frac{km}{hr}}=\frac{200}{9}\, \mathrm{\frac{m}{s}}$ denote the speeds of the chicken and the cars, respectively.

Consider the base case in which the chicken crosses perpendicularly to the road (i.e. $\theta=0$ ). The chicken will have the shortest distance to travel, but she won't be moving away from the oncoming cars in the process.

If $\theta<0$ , she will have to travel a longer distance and will be moving closer to the oncoming cars at the same time, so obviously this will increase her probability of being hit.

If $\theta>0$ , then she will be travelling a longer distance, but will be moving away from the oncoming cars at the same time. Her probability of being hit will be lowest when the difference between the time it takes her to travel the extra distance compared to the base case and the time it takes a car to travel the extra North-South distance $h$ is minimized, so that the chicken gains more distance in the North-South direction than the car can travel in the extra time. That is, when $$\frac{L-w}{v 1}-\frac{h}{v 2}$$ is minimized.

Manipulating this expression yields

$$\frac{w\sec\theta - w}{v 1}-\frac{w\tan\theta}{v 2}$$

$$\frac{1-\cos\theta}{v 1}-\frac{\sin\theta}{v 2}$$

To find when it achieves its minimum, we take its derivative with respect to $\theta$ and set it equal to zero.

$$\frac{\sin\theta}{v 1}-\frac{\cos\theta}{v 2}=0$$

$$v 2 \sin\theta - v 1 \cos\theta=0$$

$$\tan\theta=\frac{v 1}{v 2}$$

$$\theta=\tan^{-1}\left(\frac{v 1}{v 2}\right) = \tan^{-1}\left({\frac{.25\,\mathrm{\frac{m}{s}}}{200/9 \, \mathrm{\frac{m}{s}}}}\right)$$

$$ =\tan^{-1}\left(\frac{9}{800}\right) = .01125.$$

To verify that this is a minimum, we take the second derivative of our original expression and evaluate it at $.01125$ .

$$\left[{\frac{\cos\theta}{v 1}+\frac{\sin\theta}{v 2}}\right]_{\theta=.01125}.$$

This is clearly positive for all $0<\theta<\frac{\pi}{2}$ because the sine and cosine functions are both positive for these values of $\theta$ . Because the second derivative is positive at $\theta=.01125$ , we have verified that this is a minimum, which proves that this value of $\theta$ yields the maximum probability of survival for the chicken.

First we put the referencial frame on the cars, so the resulting vector from the velocity sum will change accordingly to the direction of the chicken's velocity vector.

Note that we can make this sum on the following way:

draw the 80Km vector and then add the chicken vector on its top, but since the direction is arbitrary, the choices will make a semicircunference of points. The one that makes the smallest vector sum i.e. the smallest distance (yeldind the smallest time since the speed is constant) is when this vector sum is tangent to the semicircunference.

After some geometry, we get that the angle is asin(Vch/Vca)

Vch = velocity of the chicken Vca = velocity of the cars

asin(9/8000) = 0.01125

Let's let the chicken start across the road with speed $v_c$ at time $t=0$ and start at the origin of our coordinate system. The $x$ and $y$ positions of the chicken are given by

$\begin{aligned} x_c &= t v_c \cos \theta \\ y_c &= t v_c \sin \theta. \end{aligned}$

If $W$ is the width of the road, the time the chicken gets across is given by

$t_a=\frac{W}{v_c \cos \theta}.$

The $y$ position of the next car that might hit the chicken is

$y_n=v_n t - y_0$

where $v_n$ is the speed of the car and $y_0$ is its starting point. When the $y$ position of the chicken and the car are equal, the chicken will get hit. Therefore

$y_n=y_c \rightarrow v_n t -y_0=v_c t \sin \theta\rightarrow t_h=\frac{y_0}{v_n-v_c \sin \theta}$

where $t_h$ is the time at which the chicken gets hit. The chicken does not get hit if $t_a<t_h$ . Therefore the chicken has the best chance of getting across the road when $t_a/t_h$ is a minimum, and so we have

$\frac {d(t_a/t_h)}{d \theta}=0$

which allows us to solve for $\sin \theta=v_c/v_n=0.01125$ and so $\theta=0.011~\mbox{radians}$ .

For the chicken to travel safely across the road, the time the chicken takes must be less than the time a car takes to reach the chicken.

Let's call w the width of the road and D the initial distance from the incoming car to the chicken.

Let's also say that the chicken crosses the road at an angle $\theta$ north of east. We can then say that the chicken's horizontal speed is $0.25 \cos \theta$ and that its vertical speed is $0.25 \sin \theta$ . Since D has a log-normal distribution, $\log D$ is normally distributed.

You can draw this problem as vector: Car from north to south,
Chicken (0.25 m/s) \ | \ | a \ | South _ _ _ _ \|North, Car (80 km/hours)

We get angle $\cos a=\frac {0.25}{{\frac{80000}{3600}}}$ Angle of chicken from north of east $\theta=\frac {\pi}{2}-\arccos a=0.01125$