Show that the distance between the bodies after t seconds

Key takeaway: To show the distance formula, find each position vector at time $t$, subtract to obtain the relative position, and then simplify the squared distance.

Step 1: Position vectors after $t$ seconds

For $P$:

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

For $Q$:

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

Step 2: Form the separation vector

Using $\mathbf{PQ}=\mathbf{r}_Q-\mathbf{r}_P$,

$\mathbf{PQ}=(-3+t-(5+2t),\,13-(7-2t))$

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

Step 3: Find the distance

The distance is the magnitude of $\mathbf{PQ}$:

$|\mathbf{PQ}|=\sqrt{(-8-t)^2+(6+2t)^2}$

Expand the squares:

$(-8-t)^2=(t+8)^2=t^2+16t+64$

$(6+2t)^2=4t^2+24t+36$

Add them:

$|\mathbf{PQ}|=\sqrt{t^2+16t+64+4t^2+24t+36}$

$|\mathbf{PQ}|=\sqrt{5t^2+40t+100}$

Result

$\boxed{\text{Distance after }t\text{ seconds }=\sqrt{5t^2+40t+100}}$

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Pitfalls the pros know 👇 Do not square the components before subtracting the vectors. The correct order is: position of $Q$ minus position of $P$, then square the components of the relative vector.

What if the problem changes? You can also write the relative position as $\mathbf{r}_P-\mathbf{r}_Q=(8+t,\,-6-2t)$. The distance is the same because magnitude does not change when the vector direction is reversed.

Tags: relative position vector, magnitude, constant velocity