2
$\begingroup$

I'm having a hard time solving this problem concerning relative motion:

A hare runs with a constant speed $u$ on a straight line. A dog chases the hare by always running straight to it with speed $v$. Calculate the acceleration vector of the dog and its magnitude at the moment in which his velocity vector is perpendicular to the velocity vector of the hare. Suppose that at that instant the distance from the dog to the hare is equal to $d$.

So at this instant the situation looks like this:

enter image description here

Suppose the hare runs to the right. I tried solving this with relative velocities ($O$ is the origin and subscript $xy$ means $x$ relative to $y$):

$$\vec{v}_{DO} = \vec{v}_{DH} + \vec{v}_{HO}$$

But here I seem to get stuck because I don't have any clue what $\vec{v}_{DH}$ could be. There seems to be a paradox in my brain because this vectors direction changes constantly because the position of the hare and the dog changes. But the position of the dog depends on the velocity vector of the dog and that I am supposed to find ! Is there something I am missing in connection with the concept of relative motion ?

Note: I haven't had an in depth differential equations course yet, but I can solve simple ones.

$\endgroup$
2
$\begingroup$

Suppose the dog has coordinates $\vec{r}_D(t) $, and the hare, $\vec{r}_H(t)$. Defining $\vec{r}_{HD} =\vec{r}_H-\vec{r}_D $, we can conclude $\dot{\vec{r}}_D(t)=v\hat{r}_{HD}$, where $\hat{r}_{HD}\equiv\frac{\vec{r}_{HD}}{|\vec{r}_{HD}|}$. We now simply take a time derivative of this velocity to get the acceleration. Note, we need the product rule, however, $v$ is a constant so we're left with $\ddot{\vec{r}}_D(t)=v\dot{\hat{r}}_{HD}$. This form is independent of the choice of coordinates. Given the hare's motion, you know both and may choose an ideal coordinate system and then differentiate the unit vector for a specific case.

$\endgroup$
  • $\begingroup$ The magnitude of the dog's velocity is $v$, not $u$. $\endgroup$ – Aaron Stevens Aug 14 '18 at 4:03
  • $\begingroup$ Also, you don't want to set the length of the vector to d until after you take the derivative. $\endgroup$ – Aaron Stevens Aug 14 '18 at 20:58
  • $\begingroup$ @Aaron, but it's a unit vector not of length $d$ $\endgroup$ – Captain Morgan Aug 14 '18 at 21:00
  • $\begingroup$ But you are assuming they are a distance $d$ apart to determine this. This assumption cannot be made so early on. This is essentially a related rates problem (like in introductory calculus). You have to find an equation for the derivative in general first, then find what the derivative is when they are a distance $d$ apart with perpendicular velocities. You can't put in the assumptions before finding the derivative. $\endgroup$ – Aaron Stevens Aug 14 '18 at 23:34
  • $\begingroup$ Oh yes indeed, I was mislead by the picture. Yeah just normalize later $\endgroup$ – Captain Morgan Aug 14 '18 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.