Small bodies P and Q are moving with constant velocities

Key concept: This is a standard relative-position vectors question. The key idea is to write the position of each body after time $t$, then subtract the vectors to get the separation vector and its magnitude.

Step by step

Step 1: Write the position vectors after 2 seconds

For $P$:

$\mathbf{r}_P=(5,7)+2(2,-2)=(5+4,\,7-4)=(9,3)$

For $Q$:

$\mathbf{r}_Q=(-3,13)+2(1,0)=(-3+2,\,13)=(-1,13)$

Step 2: Find the separation vector

$\mathbf{PQ}=\mathbf{r}_Q-\mathbf{r}_P=(-1-9,\,13-3)=(-10,10)$

Step 3: Find the distance

$|\mathbf{PQ}|=\sqrt{(-10)^2+10^2}=\sqrt{100+100}=\sqrt{200}=10\sqrt{2}$

Answer

The distance between the bodies after two seconds is

$\boxed{10\sqrt{2}\text{ m}}$

Pitfall alert

A common mistake is to subtract the initial positions only and forget to include the displacement from the velocities. Another error is reversing the subtraction; either order is fine for distance, but the final magnitude must be taken correctly.

Try different conditions

If the time were $t$ seconds instead of 2, you would first form

$\mathbf{r}_P=(5+2t,\,7-2t),\qquad \mathbf{r}_Q=(-3+t,\,13)$

Then the separation vector becomes

$\mathbf{PQ}=(-8-t,\,6+2t)$

and the distance is $\sqrt{(-8-t)^2+(6+2t)^2}$.

Further reading

position vector, relative velocity, magnitude of a vector