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November 3rd, 2019, 08:21 PM  #1 
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  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?. 
November 3rd, 2019, 08:29 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,647 Thanks: 1476 
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 
November 4th, 2019, 05:09 PM  #3  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Can somebody check this? Quote:
For the camper, that would be: $v_{c}=\left\langle 8016t , 0 \right \rangle$ 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: $v_{c}+v_{b}=\left\langle 8016t , 80+12t \right \rangle$ Thus the norm would be: $\left\v_{c}+v_{b}\right\=\sqrt{\left(8016t\right)^2 +\left(80+12t \right )^2}$ $\left\v_{c}+v_{b}\right\=\sqrt{80\left(5t^256t+160\right)}$ 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: $\frac{4\left(5t28\right)}{\sqrt{t^2\frac{56t}{5}+32}}=0$ 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). $4\left(5t28\right)=0$ 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.  
November 4th, 2019, 05:16 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,647 Thanks: 1476 
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. 
November 4th, 2019, 07:20 PM  #5  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
This would become into: $160\left(5t28\right)=0$ 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.  
November 4th, 2019, 08:26 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,647 Thanks: 1476 
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. 
November 4th, 2019, 11:56 PM  #7  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
 

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