Is there a way to simplify finding the minimum distance between two objects approach?

Jun 2017
218
6
Lima, Peru
The problem is as follows:

Two buggies one orange and the other blue are passing through an intersection with constants speeds of $v_{orange}=6\,\frac{m}{s}$ and $v_{blue}=6\sqrt{3}\,\frac{m}{s}$ respectively. Find the minimum approach in meters between the orange and blue buggies.​



The alternatives given in my book are:

$\begin{array}{ll}
1.&10\,m\\
2.&12\,m\\
3.&24\,m\\
4.&36\,m\\
\end{array}$

What I attempted to do is to establish the equations in the intersection as follows:

For x:

$-40\cos23^{\circ}+6t$

For y:

$-40\sin23^{\circ}+6\sqrt{3}t$

Therefore the distance for those would be the norm.

$\left\langle -40\cos23^{\circ}+6t, -40\sin23^{\circ}+6\sqrt{3}t \right\rangle$

$\left\|\left\langle -40\cos23^{\circ}+6t, -40\sin23^{\circ}+6\sqrt{3}t \right\rangle\right\|=d$

$d^2=\left(-40\cos23^{\circ}+6t\right)^2+ (-40\sin23^{\circ}+6\sqrt{3}t)^2$

Taking the derivative and equating to zero would give me the minimum time isn't it?

$\left(\left(-40\cos23^{\circ}+6t\right)^2+ (-40\sin23^{\circ}+6\sqrt{3}t)^2\right)'=(d^2)'$

$(d^2)'=12\left(-40\cos23^{\circ}+6t\right)+12\sqrt{3}\left(-40\sin23^{\circ}+6\sqrt{3}t\right)=0$

Simplifying:

$-40\cos23^{\circ}+6t+\sqrt{3}\left(-40\sin23^{\circ}+6\sqrt{3}t\right)=0$

$-40\cos23^{\circ}-40\sqrt{3}\sin23^{\circ}=-18t-6t$

$-40\cos23^{\circ}-40\sqrt{3}\sin23^{\circ}=-24t$

Further simplification and multiplication by $(-1)$:

$5\cos23^{\circ}+5\sqrt{3}\sin23^{\circ}=3t$

$t=\frac{5\left(\cos23^{\circ}+\sqrt{3}\sin23^{\circ}\right)}{3}$

Then condensing the trigonometric expressions by:

$t=\frac{5\times 2\left(\frac{1}{2}\cos23^{\circ}+\frac{\sqrt{3}}{2}\sin23^{\circ}\right)}{3}$

$t=\frac{5\times 2\left(\sin 30^{\circ}\cos23^{\circ}+\cos 30^{\circ}\sin23^{\circ}\right)}{3}$

$t=\frac{5\times 2 \sin 53^{\circ}}{3}$

Assuming that $\sin 53^{\circ}= \frac{4}{5}$

$t=\frac{5\times 2 \times \frac{4}{5}}{3}=\frac{8}{3}$

Then this would be the minimum time for both buggies to be closer.

Therefore the minimum distance between them would be the square root of evaluating that time in the function stated in the beginning:

$d^2=\left(-40\cos23^{\circ}+6t\right)^2+ (-40\sin23^{\circ}+6\sqrt{3}t)^2$

But here's where it is coming ugly:

$d^2= 40^2\cos^2 23^{\circ}-480\cos23^{\circ}t+36t^2$

$d^2=40^2\cos^{2}23^{\circ}-480\cos23^{\circ}t+36t^2+40^2\sin^{2}23^{\circ}-480\sqrt{3}\sin23^{\circ}t+108t^2$

$d^2=40^2\sin^{2}23^{\circ}+40^2\cos^{2}23^{\circ}-480t\left(\cos23^{\circ}+\sqrt{3}\sin23^{\circ} \right )+144t^2$

$d^2=40^2\left(\sin^{2}23^{\circ}+\cos^{2}23^{\circ}\right )-480\times 2t \left(\frac{1}{2}\cos23^{\circ}+\frac{\sqrt{3}}{2}\sin23^{\circ} \right )+144t^2$

$d^2=40^2\left(1\right )-480\times 2t \left(\sin30^{\circ}\cos23^{\circ}+\cos30^{\circ}\sin23^{\circ} \right )+144t^2$

$d^2=1600-480\times 2t \left(\sin53^{\circ} \right )+144t^2$

Again assuming $\sin 53^{\circ}= \frac{4}{5}$:

$d^2=1600-960t\left(\frac{4}{5} \right)+144t^2$

$d^2=1600-768 t+144t^2$

Then all will left to do is taking the square root to that expression evaluating with $t=\frac{8}{3}$

$d=\sqrt{1600-768 \left(\frac{8}{3}\right)+144\left( \frac{8}{3} \right)^2}$

$d=\sqrt{1600-768 \left(\frac{8}{3}\right)+144 \left( \frac{64}{9}\right) }$

$d=24$

Therefore the minimum distance would be $24\,m$

I believe this would be the answer but as it can be seen, this procedure was very exhausting in terms of calculation and keeping up with the right numbers. Does it exist a way to simplify it or to ease it?. Can somebody help me here?. :help: :eek:
 
Last edited:
Jun 2019
439
236
USA
They always make a right triangle, with their distance as the hypotenuse, don't they?
Does that help?