Calculate time to collision? I have two objects ($a$ and $b$) moving towards each other at different speeds ($30$mph and $50mph$ respectively). The distance ($d$) between them is, let's say 2 miles.
I initially came across Kinematics in this post, but it led to incorrect results. I focused on the simple statement:

Simpler still - look at the difference in velocity. If one goes at 25 km/h and the other goes at 50 km/h, the faster one is catching up on the slower one at a speed of (50-25)=25 km/h. So whatever the gap between them at the start, that's the gap that he is closing at that speed.
Then the time taken to close the gap is (initial gap) / (speed of closing the gap), and once you have the time, you can calculate the distance traveled because you have the speed.

This gave me a calculation of:
$$\frac{2mi}{50mph - 30mph} = \frac{2mi}{20mph} = 0.1$$
Multiplying this by 60 minutes gave me an output of $6$ minutes to close the gap, which seemed reasonable. However, when scaling it down $a = 2, b = 3, d = 0.5$, the answer doesn't seem correct anymore:
$$\frac{0.5mi}{3mph - 2mph} = \frac{0.5mi}{1mph} = 0.5$$
Multiplying this by 60 minutes gives me $30$ minutes to close the gap which tells me that I'm either misunderstanding something about the solution, performing a miscalculation, or this isn't the solution I need.

How do I calculate the amount of time, in minutes it takes for the two objects to close this gap and collide, neglecting any additional forces?

Note: This question relates to a puzzle, not school work. It is purely recreational, but I prefer to learn the solution, not have it simply given to me. I want to understand it.
 A: When two objects are moving towards each other, the sum of their velocities is needed for this calculation, not the difference.
Special thanks to @lulu for helping me to understand the problem, and the solution.

Defining Variables
For the entirety of this answer, let the distance $d$, between two objects $a$ and $b$, remain at a constant value of $2$ miles. The variables $a$ and $b$ will represent the velocities of the aforementioned objects respectively.
Colliding With a Stationary Object
For the first example, assume that $a$ is moving towards $b$ at a rate of $30$mph and $b$ is not moving at all. The time (in minutes) it takes $a$ to collide with $b$ is easily calculated as:
$$t = \frac{d}{a} \cdot 60$$
So, $2$ miles, divided by a speed of $30$ miles per hour, multipled the $60$ minutes in an hour, gives us a travel time of $4$ minutes.
Head-On Collision
For this example, assume that $a$ and $b$ are moving towards each other:

The time to collision can be calculated in the same manner as before, only we have to add the velocities of $a$ and $b$ together:
$$t = \frac{d}{a + b} \cdot 60$$
So, assuming that $a$ is moving at $30$mph and $b$ is moving at $50$mph; the distance of $2$ miles will be covered at a rate of $80$mph. As such, $2$ miles, divided by $80$mph multiplied by the 60 minutes in an hour, gives a collision time of $1.5$ minutes.
Passing Collision
Finally, in this example, assume that $a$ and $b$ are moving in the same direction:

The time to collision can be calculated in a similar manner to a head-on collision. Simply use the difference between the velocities instead of their sum:
$$t = \frac{d}{\lvert a - b\rvert} \cdot 60$$
So, assuming that $a$ is moving at $30$mph and $b$ is moving at $50$mph; the distance of $2$ miles will be covered at a rate of $20$mph. As such, $2$ miles, divided by $20$mph multiplied by the 60 minutes in an hour, gives a collision time of $6$ minutes.
A: HINTS:

Solve for $x1,x2$
$$ x1+ x2==L =2 \tag 1  $$
$v1,v2$ are speeds of two objects.
$$ x1/ v1 = x2/v2 \;( \;= T); \quad x1+x2= L. \tag 2 $$
and find $T$.
Solving
$$ x1=0.75, x2= 1.25 \text{ miles}$$
now you can find T.
A: Let $x$ be the distance traveled by the first object, and $y$ the distance traveled by the second one. When they meet and collide, both objects will have covered a total distance of $2$ mi, so $x+y=2$. At that time, you have $y=2-x$.
The time it takes for the first object to travel $x$ mi is $\frac{x\,\rm mi}{30\frac{\rm mi}{\rm h}}=\frac x{30}$ h or $2x$ min.
The time it takes for the second object to travel $y=2-x$ mi is $\frac{2-x\,\rm mi}{50\frac{\rm mi}{\rm h}}=\frac{2-x}{50}$ h or $\frac{12-6x}5$ min.
These times must be the same, so
$$2x = \frac{12-6x}5 \implies x=\frac34$$
which means the objects collide after $2\cdot\frac34=\frac32=1.5$ min, when the objects have traveled respective distances of $0.75$ mi and $1.25$ mi.
