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November 3rd, 2019, 08:21 PM   #1
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Math Focus: Calculus
Question How can I find the time for the least possible distance between two cars?

The problem is as follows:

Two volkswagen possesses constant velocities and pass through the position labeled $A$ on the instant indicated in the figure from below. Find the time which will take the cars to be separated by a minimum distance.



The alternatives given on my book are as follows:

$\begin{array}{ll}
1.4\,s\\
2.5.6\,s\\
3.8\,s\\
4.9.6\,s\\
5.12\\
\end{array}$

This problem seems to be related to the use of differentials but I don't know exactly on what way should I use them.

From the given data I could only spot these relationships.
Assuming $v_{b}=\textrm{beetle}$ and $v_{c}=\textrm{camper}$

$v_{b}=0_{i}\hat{i}+12t\hat{j}$
$v_{c}=-16t\hat{i}+0\hat{j}$

But that's it. Then what?. Can someone with experience help me what should I do next?. How should I find the time when the distance that separates them (after crossing the intersection I believe) is the least possible?.
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November 3rd, 2019, 08:29 PM   #2
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find the position of both cars as a function of time

then find the squared distance between them as a function of time

minimize that using calculus or by finding the vertex of the resulting parabola
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November 4th, 2019, 05:09 PM   #3
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Question Can somebody check this?

Quote:
Originally Posted by romsek View Post
find the position of both cars as a function of time

then find the squared distance between them as a function of time

minimize that using calculus or by finding the vertex of the resulting parabola
I attempted to do that in what I wrote in my question. Assuming that the observer is placed in the intersection of the two vehicles.

For the camper, that would be:

$v_{c}=\left\langle 80-16t , 0 \right \rangle$

For the beetle, would be:

$v_{b}=\left\langle 0 , -80+12t \right \rangle$

I'm assuming that for the person who's standing in the intersection the camper has a negative velocity as is going to the left and the beetle is having a positive "vertical" velocity as it is moving towards him. But am I right with this conclusion?

Then the sum of both vectors would be the resultant. I'm still not sure why I'm doing this. Just is my attempt to follow your directions as a cookbook.

Then the sum is:

$v_{c}+v_{b}=\left\langle 80-16t , -80+12t \right \rangle$

Thus the norm would be:

$\left\|v_{c}+v_{b}\right\|=\sqrt{\left(80-16t\right)^2 +\left(-80+12t \right )^2}$

$\left\|v_{c}+v_{b}\right\|=\sqrt{80\left(5t^2-56t+160\right)}$

Thus if what is asked is to minimize this then it would imply that I should differentiate and equate to zero.

But this looks kind of too long algebraically.

Well after doing this procedure I obtained:

$\frac{4\left(5t-28\right)}{\sqrt{t^2-\frac{56t}{5}+32}}=0$

Then the next step would be equate to zero the numerator: (Assuming what it comes ahead discards the solutions for $t$ when the denominator is zero).

$4\left(5t-28\right)=0$

Thus:

$5t=28$

and

$t=\frac{28}{5}=5.6\,s$

Which correspond to the second answer and it checks with the right answer. But again. Why on earth I am doing this? What's the reason behind this whole thing? Why a sum? Can somebody help me with the conceptual explanation please?

Last edited by skipjack; November 4th, 2019 at 10:42 PM.
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November 4th, 2019, 05:16 PM   #4
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try it again but use the squared norm.

minimizing the squared distance also minimizes the distance.

there was a reason I said to use the squared distance and you found it the hard way.
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November 4th, 2019, 07:20 PM   #5
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Originally Posted by romsek View Post
try it again but use the squared norm.

minimizing the squared distance also minimizes the distance.

There was a reason I said to use the squared distance and you found it the hard way.
Okay If I do that...

This would become into:

$160\left(5t-28\right)=0$

Then:

$t=\frac{28}{5}=5.6\,s$

Then this corresponds to the answer as indicated. But there is still pending the answer to the question which I posted above. Why are we looking for the resultant of these two velocities? Is it because that for an observer who's in the intersection? To him, the velocity will be the resultant? And, is my assumption regarding the signs of the velocities as I assigned them to the camper and the beetle, correct? Why does minimizing the squared norm also work as does minimizing the norm?

Last edited by skipjack; November 4th, 2019 at 10:40 PM.
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November 4th, 2019, 08:26 PM   #6
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Your question confuses me.

We are given the initial positions and constant velocities of the two vehicles.

Thus we are able to formulate their positions, and thus their distance, as a function of time.

We then minimize this using whatever means. You arrived at the correct answer.
At 5.6s seconds the vehicles are as close as they get.

That's all there is to it. There's no observer at the intersection... It's true that we set the intersection as the (0,0) point of the coordinate system we use for the problem but there is no observer there.
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November 4th, 2019, 11:56 PM   #7
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Quote:
Originally Posted by romsek View Post
Your question confuses me.

We are given the initial positions and constant velocities of the two vehicles.

Thus we are able to formulate their positions, and thus their distance, as a function of time.

We then minimize this using whatever means. You arrived at the correct answer.
At 5.6s seconds the vehicles are as close as they get.

That's all there is to it. There's no observer at the intersection... It's true that we set the intersection as the (0,0) point of the coordinate system we use for the problem but there is no observer there.
When I read this problem I thought that there's a reference frame or something. My question was referring to why should I add both positions to get a norm? is it because the only way to obtain the least possible distance between them can only be established by the diagonal which separates them?.
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