Probability of intersecting paths on loop under random starting condition Jim leaves his house at noon and walks at a constant rate of 4 miles per hour along a 4 mile loop (returning to his house at 1 PM). At a randomly chosen time, uniform between noon and 1 PM, Sally chooses a random location (uniform along Jim’s route) and begins running at a constant rate of 7 miles per hour along the route in the same direction that Jim is walking until she completes one 4-mile circuit of the route.
The probability that Sally runs past Jim while he is walking is given by $m/n$, where $m$ and $n$ are relatively prime integers. 
Find $m + n$.

I find this problem interesting because of the condition that Sally completes just a single loop and cannot continue, even if it would allow her to catch Jim.  This makes finding the appropriate starting conditions a bit tricky.  Moreover, it is surprising to me that the answer will be the ratio of two primes.  This is certainly not obvious from the problem statement.
 A: Jim's trajectory is in blue.  If and only if Sally starts anywhere in the shaded regions, she will catch Jim.
Note to all:  be careful to include the fact that Sally can run just once around the loop... she cannot go farther, even if it would allow her to catch Jim!

Of course negative distances are simply measured backward from Jim's house.
The shaded area is the sum of two triangles:
$$A = \underbrace{\frac{1}{2} \cdot \frac{12}{7} \cdot \frac{3}{7}}_{orange} + \underbrace{\frac{1}{2} \cdot \frac{3}{7} \cdot 4}_{blue} $$
and the total possible area is $1 \cdot 4 = 4$, so the ratio is $15/49$ and thus the sum is $64$.

Note this is the same answer as David K:
$$\int\limits_{t=0}^1 \int\limits_{D=0}^{Min[3 t, 12/7]} 1\ dt\ dD = \frac{15}{49}$$
A: The two end points of this distribution spread over the $60$ minute time frame will be $(0,.75)$ and $(60,0)$ where $x$ is the start time of Sally in minutes and $y$ the probability of Sally passing Jim. At $t = 0$, the probability of Sally passing Jim is the probability of Sally starting no more than $3$ miles behind Jim on the course. This gives a position of any $3$ miles out of $4$ hence $p = 0.75$. At $t = 60$, it is impossible for Sally to pass Jim hence $p = 0$. All you have to do now is determine the shape of the distribution between these two end points (try $t = 30$) and integrate to get the overall probability. Convert it to a fraction and you are done.
Revision $2$ Edit:
Oops, Sally only completes one $4$ mile circuit so the revised probability distribution will be:

Revision 3 addition
Another way to look at the probability distribution is to plot p versus starting position of Sally in miles from Jim's house. The significance of $s = 2 \frac{2}{7}$ is that it's the furthest distance from Jim's House along the course direction whereby Sally won't pass Jim if she leaves at noon. The area divided by $4$ gives the probability.

A: At the time and place where Sally starts, Jim is $D$ miles ahead and will continue walking for $T$ hours.
Sally will run for $\frac47$ hour. Since she gains $3$ miles per hour on Jim, if $D > \frac{12}{7}$ she will not pass him before she stops running.
If $D > 3T$ then Sally will not pass Jim before he stops walking.
But if $D < 3T$ and $D < \frac{12}{7}$, Sally will pass Jim.
(I'm ignoring the cases $D = 3T$ and $D = \frac{12}{7}$ because they have zero probability and I don't care to quibble about the definition of "passing" in those cases.)
The joint distribution of $D$ and $T$ is uniform over the rectangle
$0 \leq D < 4$ and $0 \leq T \leq 1.$
Figure out the fraction of this region that lies under both of the lines
$D < 3T$ and $D < \frac{12}{7}$.
