How can I find the time for the least possible distance between two cars moving to intersect each other?

Question

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?.

Answer 1

First of all, notice that both vehicles are 80m away from point $A$ and they have different speeds, so they are not going to crash (this implies the minimum distance is not zero). Now, as time goes by, both vehicles move towards point $A$. You can express the distance $l(t)$ between them as the hypothenuse of the right triangle of sides $80 - v_{b}t = 80-12t$ and $80 - v_{c}t = 80-16t$. Thus, the distance is $$l(t) = \sqrt{(80-12t)^{2}+(80-16t)^{2}}$$ Finally, if you want to find the minimum distance, you want to find $t$ so $l(t)$ is minimum, so it is enough to solve $$\frac{dl(t)}{dt} = 0$$

Answer 2

Distance between cars = $\sqrt{\text{difference of horizontal positions of cars}^2 + \text{difference of veritical position of cars}^2}$

So we need to figure the horizontal and distances of each car.

Horizontal position of the bug and Vertical position of the bus are constant at $0$. (I'll assume $A = (0,0)$ is the origin.)

Horizontal position of the bus is $80 - 16t$. Vertical position (I'll assume down is a positive value...because it doesn't make any practical difference) is $80 - 12t$.

So

$D(t)=$Distance between cars = $\sqrt{\text{difference of horizontal positions of cars}^2 + \text{difference of veritical position of cars}^2}$

$=\sqrt{(\text {horizontal position of bus}-\text{horizontal position of bug})^2+(\text {vertical position of bus}-\text{vertical position of bug})^2}$

$=\sqrt{([80-16t]-0)^2 + (0-[80-12t]^2)}$.

$=4\sqrt{(20-4t)^2 + (20-3t)^2}$

$= 4\sqrt{400-160t + 16^2 + 400 - 120t + 9t^2}$

$= 4\sqrt{800 -280t +25t^2}$

So $D'(t) = 4*\frac 12*\frac 1{\sqrt{800-280t+25t^2}}(50t -280)$

And $D'(t) = 0$ means $50t -280=0$ so $t = \frac {28}{5} = 5.6$ second.

Could we have down that without all that work?

Well we know by chain rule that $[f(g(x))]' = f'(g(x))*g'(x)$ and so $[f(g(x))]'=0$ means either $f'(g(x)) = 0$ or $g'(x)$

so $D(t) = \sqrt{stuff}$ then $D'(t) = \frac 12\frac 1{\sqrt{stuff}}*stuff'$ and as $\frac 1{morestuff}\ne 0$ so we just have to solve $stuff' = 0$.

So $[(80-16t)^2 +(80-12t)^2]' = -2(80-16t)16-2(80-12t)12 =0\implies$

$(20-4t)4+(20-3t)3=0\implies$

$25t = 140$

$t=\frac{140}{25}=5.6$

Hm... that wasn't any easier....