# 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.

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.

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.

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.

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.

• As it currently stands, this is a very confusing answer, to the point of almost being considered as "not an answer" but instead a comment. Can you take some time to add some details for future readers and explain how it relates to the question? I understand that $0.5$ means half an hour, and that's why it didn't make sense. Sep 28, 2021 at 17:53
• OK shall soon delete if not useful. Sep 28, 2021 at 18:00
• Well I'm not understanding what $L$, $x1$ and $x2$ are here. I was able to infer $T$, $v1$ and $v2$. I think if it was edited to include more information and related to the question directly (e.g. using one of the examples) it would be more useful to me (and potentially future readers). Sep 28, 2021 at 18:02