# Calculate time to collision

algebra-precalculuskinematics

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.