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So, here is the question: Assuming the 2D Cartesian system and the basis vectors $\hat{i}$ and $\hat{j}$, we have two cars A and B located at (0,6) and (-30,0) respectively. Car A starts moving with a constant velocity of -3 m/s $\hat{j}$ and Car B with 4 m/s $\hat{i}$. The question is to find the minimum separation between these two cars.

I understand there is a way we could solve this using calculus (specifically derivatives), but that way involves a lot of calculations. So my professor explained a way to solve this using relative motion, i.e assuming either body's frame and working out the problem in that frame. In here, particularly, if we assume to be in the frame of B, we would see that A has a velocity of $-(4 \hat{i}+3 \hat{j}) \frac{m}{s}$. If we draw out the line that the body A traverses in the frame of B, we get the line 3x-4y+24=0. Now the thing which the professor did was confusing to me, they said that the minimum separation between them is going to be the shortest distance of the point where car B was initially, to the line I mentioned above. What I do not get is that, why did they say that the shortest distance is from that point (-30,0) to the line? I mean it felt weird to me because B's origin (their own particular co-ordinate system's) is changing all over the time. So how did they conclude we have to calculate the distance from there? I am sorry if I could not make complete sense.

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  • $\begingroup$ the minimum separation between those two cars is zero because those two cars are collide at t=5.14 [s] $\endgroup$
    – Eli
    Oct 26, 2021 at 15:33
  • $\begingroup$ @Eli, it's not possible, because at t=5.14 s, car A is at (0, -9.42) and the car B is at (-9.44, 0). The distance between them is not zero. $\endgroup$
    – Floatoss
    Oct 26, 2021 at 15:47
  • $\begingroup$ from your data I got those two equations car A $y=6-3\,t~$ car B $~x=-30+4\,t$ $\endgroup$
    – Eli
    Oct 26, 2021 at 15:53
  • $\begingroup$ Yes, how did you conclude that they collide then? $\endgroup$
    – Floatoss
    Oct 26, 2021 at 16:00
  • $\begingroup$ well y=x is wrong ? $\endgroup$
    – Eli
    Oct 26, 2021 at 16:04

3 Answers 3

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The intuitive reason why you want the shortest distance from $B$ to the relative path is that if you were observing the motion of $A$ from car $B$, $A$ would appear to be moving along this relative path and would miss you by a certain distance, i.e. the shortest distance, because you (now located at the origin) are not on this path.

It probably doesn't help that the relative path you have quoted in your question is not actually correct:

The velocity of $A$ relative to $B$ is $$\left(\begin{matrix}0\\-3\end{matrix}\right)-\left(\begin{matrix}4\\0\end{matrix}\right)=\left(\begin{matrix}-4\\-3\end{matrix}\right)$$ The initial position of $A$ relative to $B$ is $$\left(\begin{matrix}0\\6\end{matrix}\right)-\left(\begin{matrix}-30\\0\end{matrix}\right)=\left(\begin{matrix}30\\6\end{matrix}\right)$$ Therefore at time $t$ the position of $A$ relative to $B$ is $$\underline{r}=\left(\begin{matrix}30\\6\end{matrix}\right)+t\left(\begin{matrix}-4\\-3\end{matrix}\right)$$

Note that the origin does not satisfy this equation, so there will be no collision.

In Cartesian form, this comes from $x=30-4t$ and $y=6-3t$. and eliminating $t$ gives $$\frac{30-x}{4}=\frac{6-y}{3}\implies 3x-4y-66=0$$

So the shortest distance between the cars is $\frac{66}{5}$

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  • $\begingroup$ The initial position for (A) is (6,0). Then x = 36-4t and y = 0 -3t. $\endgroup$
    – R.W. Bird
    Oct 26, 2021 at 18:50
  • $\begingroup$ @R.W.Bird that's not what the question says $\endgroup$ Oct 26, 2021 at 21:41
  • $\begingroup$ It looks to me like the instructor mapped "the trajectory of A in the frame of B" on the coordinate space of the non-frameshifted origin, with predictably confusing results. The OP's line equation can indeed be used to find the distance of closest approach by drawing the perpendicular line through (-30, 0) , as the instructor describes... but only because the line isn't the trajectory of A in that coordinate space. $\endgroup$
    – g s
    Oct 27, 2021 at 0:05
  • $\begingroup$ @user3298777, thank you for your answer. The instructor did not change the position frames, but you did. In my case, if you use the distance from (-30,0) to the line I mentioned, you get the same 66/5 as the answer. But that wasn't my question. What I am confused with is that, as B is moving (considering we don't change the position frames), how do we "fix" the coordinate of B as (-30,0) and calculate the distance from there? I mean, B is moving wrt the origin (0,0). $\endgroup$
    – Floatoss
    Oct 27, 2021 at 3:09
  • $\begingroup$ the velocity of $A$ is adjusted to allow $B$ to remain stationary $\endgroup$ Oct 27, 2021 at 9:14
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The initial frame shift is incorrect. Try shifting into the B frame, such that A is at (30, 6) on the line $3x-4y = 66$, then find the distance from that line to the origin. In the B frame, the position of B is always the origin.

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  • $\begingroup$ Thank you for your answer @gs. I get this way of going about the question, it makes complete sense. The part which still confuses me is how the moving frame's origin is always fixed, I get that in the frame of B, they always think of themselves as the origin. But I just cannot fathom it. $\endgroup$
    – Floatoss
    Oct 27, 2021 at 3:14
  • $\begingroup$ In any problem, unless it is given, you are free to choose the fame of reference. In this case, the moving frame is chosen with (B) at the origin. $\endgroup$
    – R.W. Bird
    Oct 27, 2021 at 13:32
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I get the line of motion for (A) moving in the system which is moving with (B) as: 3x – 4y -108 = 0. A perpendicular line passing through (B) (at the origin of this system) is: y = -(4/3)x. These two lines intersect at (12.96, -17.278). The minimum distance is measured from the origin to this point.

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  • $\begingroup$ I don't get how the equation of the line makes sense, what is your origin precisely @R.W Bird? $\endgroup$
    – Floatoss
    Oct 27, 2021 at 3:16
  • $\begingroup$ My mistake. I misread the question. $\endgroup$
    – R.W. Bird
    Oct 27, 2021 at 13:38

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