How to calculate the closing speed of two objects

Question

I have the 2D positions and 2D velocity vectors of two objects. How do I calculate the closing velocity ? I want to get the closing speed.

Think of one jet tracks another jet with radar and you want to yield the closing speed, which is actually the distance-change / unit. (In this example Δmiles/h)

Answer 1

You essentially have two lines with the equations:
\$a(t)=a_0 + t*v_a\$

\$b(t)=b_0 + t*v_b\$

where \$a_0, b_0, v_a, v_b\$ are vectors with the same dimension and unit (meters for \$a_0\$ and \$b_0\$ and meters per second for \$v_a\$ and \$v_b\$).

\$a(t)\$ and \$b(t)\$ are the positions of your objects at time \$t\$ (in seconds if using the units above).

Distance over time is calculated by subtracting the positions and taking the length of the vector:
\$d(t)=\left|a(t)-b(t)\right|=\left|a_0-b_0+t*(v_a-v_b)\right|\$

This has to be differentiated to get the closing speed (The multiplication used is the dot-product): \$c(t)=d'(t)=\frac{(v_a-v_b)*((v_a-vb)*t+a_0-b_0)}{\left|(v_a-v_b)*t+a_0-b_0\right|}\$

The value of \$c(t)\$ is a scalar.

You probably want the closing speed at t = 0 (i.e. now), so it can be simplified further:

\$c(0)=d'(0)=\frac{(v_a-v_b)*(a_0-b_0)}{\left|a_0-b_0\right|}\$

Unit of the result is the same as the components of the original velocity vectors. It is negative if the objects get closer to each other and positive if the distance increases. If you want it the other way around you have to multiply by -1.

\$current closing speed=-c(0)=-\frac{(v_a-v_b)*(a_0-b_0)}{\left|a_0-b_0\right|}\$

Note that this works for all dimensions.

Every matrix-/vector-math library will support all of the necessary operations. Depending on its design this can be written as a single line of code. However you may prefer to store \$a_0 - b_0\$ in a temporary variable.

In Pseudocode this may look like this

val tmp = a.position - b.position
return -((a.velocity - b.velocity).dot(tmp)/tmp.length)

Answer 2

To find the relative velocity vector, you need to substract the vectors. The magnitude of that vector is your closing speed.

(Note that the operation inside the magnitude function of the third situation is a vectoral substraction, not a scalar one.)

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Edit:

Assume initial positions are A and B, distance is dis, velocities are v(A) and v(B).

(As you can see, objects can never collide.)

Now, if we do this:

As you can see now, the c vector is the relative velocity.

Now, we can write these equations:

Also these:

Now, here is what you want in question:

As we wrote above, this yields to this:

Which eventually yields to:

Which is our relative velocity.

Answer 3

Simple: Unlike the rest of these answers.

V1*sin(dir) - V2*sin(dir)

30*sin(090) - 40*sin(270) = 70

If you want the range, just add it in like so

Range-(30*sin(090) - 40*sin(270) = 70)