Uniform Motion in 2D - A man chasing a bus

  • #1
brotherbobby
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Homework Statement
(a) A bus is running along a highway at a speed of ##v_1=16\,\text{m/s}##. A man is at a distance of ##a=60\,\text{metres}## from the highway and ##b=400\,\text{metres}## from the bus. In what direction should the man run to reach any point of the highway at the same time as the bus or before it? The man can run at a speed of ##v_2=4\,\text{m/s}##.

(b) At what minimum speed should the man run in (a) above to be able to meet the bus and in what direction?
Relevant Equations
1. In uniform motion with some velocity ##v##, distance travelled and time taken as related as ##\Delta x=v\Delta t##
2. ##\tan\theta=\dfrac{1}{\csc\theta}## and ##\csc^2\theta=1+\cot^2\theta.##
1715371076244.png
This is how the problem(s) that I called (a) and (b) appeared in the text - as two distinct problems.



Part (a) :
First, in part (a) [or Problem 9. in the text], why ask "at the same time as the bus or before it"? Surely that should mean two different questions. After a bit of thinking, I believe it is, though not quite in the way I first imagined. I do part (a) below in two different stages ##-## which describe the closest and farthest points of meeting of the man and the bus, respectively. Anywhere in between those two extremes, the man will arrive on the highway before the bus.

1715371118193.png

(1) Closest point of meeting : I draw the diagram where the bus B meets the man M at the point C, the closest point of meeting. Anywhere before C on the line BA, the bus will arrive before the man.

If ##v_B## be the velocity of the bus and ##v_C## that of the man, we have ##\text{BC}=v_Bt_1## and ##\text{MC} = v_Mt_1##. Since the times are the same at the point of meet, ##\small{\dfrac{BC}{v_B}=\dfrac{\text{MC}}{v_M}\Rightarrow \dfrac{\text{AB}-\text{AC}}{v_B}=\dfrac{\text{MC}}{v_M}\Rightarrow \dfrac{\sqrt{b^2-a^2}-a\cot\theta_1}{v_B}=\dfrac{a\csc\theta_1}{v_M}}##, using the the Pythagorean theorem for AB and trigonometric ratios for MC and MA in terms of ##a## and ##\theta_1##.
Putting values, ##\small{\dfrac{\sqrt{400^2-60^2}-60\cot\theta_1}{16}=\dfrac{60\csc\theta_1}{4}\Rightarrow 395-60\cot\theta_1=240\csc\theta_1\quad \color{blue}{-(1)}}##. Upon squaring both sides and using ##\csc^2\theta=1+\cot^2\theta##, I obtained a quadratic in ##\cot\theta_1## which solved to ##\small{\cot\theta_1=0.98\Rightarrow\tan\theta_1=1.02\Rightarrow \boxed{\boldsymbol{\theta_1=45.6^{\circ}}}}##.
1715371166895.png


This answer could be wrong, as per the text. I paste the answer on the right, where I mark the lower value of the angle ##\theta_1## (their #\alpha##) in red ink (_____). I'd paste the authors' image of the problem after I complete (b). Their lower limit for ##\alpha## is different as my ##\theta_1##. [I'd explain the meaning of the upper limit in part (b).

1715371217824.png


Incidentally, upon putting ##\color{blue}{(1)}## as a trigonometric equation in Mathway for it to solve for me, I got a different answer. It tells me the angle ##x=0.8^{\circ}, 2.64^{\circ}##, which are both far from my answer of ##\theta_1=45.6^{\circ}##.

Request : Where am I going wrong in evaluating the solution for the equation online?


1715371261369.png

(2) Farthest point of meeting : Here the bus B meets the man M at the point D, the farthest point of meeting. Anywhere after D on the line BA, the bus will arrive before the man. The time of meeting is some ##t_2##.

Thus ##\small{t_2=\dfrac{\text{BD}}{v_B}=\dfrac{MD}{v_M}\Rightarrow \dfrac{\text{AB}+\text{AD}}{v_B}=\dfrac{\text{MB}}{v_M}\Rightarrow \dfrac{\sqrt{b^2-a^2}+a\cot\theta_2}{v_B}=\dfrac{a\csc\theta_2}{v_M}\Rightarrow \dfrac{\sqrt{400^2-60^2}+60\cot\theta_2}{16}=\dfrac{60\csc\theta_2}{4}\Rightarrow 395+60\cot\theta_2=240\csc\theta_2\quad\color{blue}{-(2)}}.## I solve it as before to get an answer of ##\boxed{\boldsymbol{\theta_2=28.3^{\circ}}}##.

1715371306764.png

This answer could be wrong too. It corresponds (roughly) to the upper limit of ##\alpha## that I underline in red ink (______).

I copy and paste the authors' working diagram below. Not only is the diagram far from clear, I am not sure how their upper limit of ##\alpha## corresponds to my ##\theta_2##.

1715371341334.png


Upon matching their answers with mine, using some triangle trigonometry, I obtain their values of ##\alpha## to be : ##\boxed{\boldsymbol{36.9^{\circ}<\alpha<143.07^{\circ}}}##. I presume these are correct approximately. Note their answers are in degrees and minutes.

Request : Is my working fine? Or am I missing something?
1715371388461.png

Part (b) : To find the minimum speed I had no doubts I had to use calculus.

The man moves along CO with minimum speed ##v_m## reaching C at a distance ##x## from A, taken to be also the origin O. Since both B and M reach C at the same time ##t##, ##\small{t=\dfrac{\text{BC}}{v_B}=\dfrac{MC}{v_m}\Rightarrow\dfrac{\text{AB}+x}{v_B}=\dfrac{\sqrt{a^2+x^2}}{v_m}\Rightarrow (\text{AB}+x)v_m=\sqrt{a^2+x^2}v_B}##.
I squared this expression to obtain an expression for ##v^2_m## : ##\small{v_m^2=\dfrac{(a^2+x^2)v_B^2}{\text{AB}^2+2\text{AB}x+x^2}}##.
I would differentiate this with respect to ##x## to obtain ##\dfrac{dv_m}{dx}## in a quadratic expression terms of ##\text{AB}, x, v_B##. Setting it to zero for minimum condition, the value of ##x=9.11\,\text{m}##. This would yield the direction of the man's motion ##\boxed{\boldsymbol{\theta = 81.4^{\circ}}}## and his speed ##\boxed{\boldsymbol{v_m=2.43\,\text{m/s}}}##.

1715371426476.png

These answers were correct. In the diagram above, ##\beta+\theta=\alpha##. The value of ##\beta =8.6^{\circ}## and they indeed give the value of ##\alpha## in the text.

However, the authors did not pursue a method as tortuous as mine where I had to avail myself of the methods of calculus. The immediately surmised that for minimum speed, the man should travel perpendicular to the initial line connecting him to the bus! Also, they argued that for minimum speed of motion for the man, the times ##t_1## and ##t_2## in parts (a) and (b) for closest and farthest meetings should both be the same! But how?

Request : How is that for minimum speed of meeting, the direction of travel by the man is normal to the line connecting him to the bus? How also do the times of meeting for near and far travels become the same for minimum speed of motion by the man?

Thank you for the trouble.
 
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  • #2
brotherbobby said:
Request : How is that for minimum speed of meeting, the direction of travel by the man is normal to the line connecting him to the bus?
To meet the bus, he needs to travel distance ##a## in the direction perpendicular to the highway no matter at what point on the highway the meeting point is. His speed will be minimum if his entire velocity is in the perpendicular direction. If it weren't, then only a component of the velocity will be in the perpendicular direction which means higher speed.
 
  • #3
kuruman said:
To meet the bus, he needs to travel distance a in the direction perpendicular to the highway no matter at what point on the highway the meeting point is.
But his motion is not perpendicular to the highway. It is perpendicular to the line joining his initial position to that of the bus, which is line MB in my diagram.
 
  • #4
brotherbobby said:
Homework Statement: (a) A bus is running along a highway at a speed of ##v_1=16\,\text{m/s}##. A man is at a distance of ##a=60\,\text{metres}## from the highway and ##b=400\,\text{metres}## from the bus. In what direction should the man run to reach any point of the highway at the same time as the bus or before it? The man can run at a speed of ##v_2=4\,\text{m/s}##.

(b) At what minimum speed should the man run in (a) above to be able to meet the bus and in what direction?
Relevant Equations: 1. In uniform motion with some velocity ##v##, distance travelled and time taken as related as ##\Delta x=v\Delta t##
2. ##\tan\theta=\dfrac{1}{\csc\theta}## and ##\csc^2\theta=1+\cot^2\theta.##

How is that for minimum speed of meeting, the direction of travel by the man is normal to the line connecting him to the bus?
The time to meet, ##MC/v_m=BC/v_B##.
##v_m=v_B*MC/BC=v_B*\sin(\beta)/\sin(\alpha)##.
To minimize ##v_m## we need to maximize ##\sin(\alpha)##, which gets maximum at ##\pi/2##.
 
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  • #5
Hill said:
The time to meet, ##MC/v_m=BC/v_B##.
##v_m=v_B*MC/BC=v_B*\sin(\beta)/\sin(\alpha)##.
To minimize ##v_m## we need to maximize ##\sin(\alpha)##, which gets maximum at ##\pi/2##.
Excellent. Many thanks.
 
  • #6
brotherbobby said:
But his motion is not perpendicular to the highway.
It is when he is moving at the minimum speed needed to catch the bus. He travels at constant speed in a straight line regardless of direction. Therefore, minimum speed means minimum distance traveled in a straight line and that's the perpendicular.
 
  • #7
kuruman said:
It is when he is moving at the minimum speed needed to catch the bus. He travels at constant speed in a straight line regardless of direction. Therefore, minimum speed means minimum distance traveled in a straight line and that's the perpendicular.
Since they meet, their times of travel are the same. Hence, as you say, ##x_{min}=v_{min}t##. So indeed the distance travelled by the man should be minimum, in which case he should move perpendicular to the highway.

But that contradicts the answer of the text. See @Hill's argument in post #4 above. I think I should illuminate it by means of a diagram.

Below, ##t## is the time when they meet.

1715375860711.png



Which tells us that he should move perpendicular to the line that joins them at the start of motion.
 
  • #8
kuruman said:
To meet the bus, he needs to travel distance ##a## in the direction perpendicular to the highway no matter at what point on the highway the meeting point is. His speed will be minimum if his entire velocity is in the perpendicular direction. If it weren't, then only a component of the velocity will be in the perpendicular direction which means higher speed.
If he aims for a point slightly further away from the starting point of the bus (by angle ##\Delta\theta##), the increase in the distance the bus has to travel (hence, the increase in time available) is of a higher order (##\sin(\Delta\theta)\approx\Delta\theta##) than the increase in distance he has to travel (##\sec(\Delta\theta)-1\approx\frac 12\Delta\theta^2##). So his minimum speed must decrease.
 
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  • #9
I will rephrase what I tried to say because it didn't come out right. Consider the situation in which the man encounters the bus after selecting an appropriate velocity for that to happen. His path is a straight line angled as shown in the diagrams above. At minimum speed, his path will be of minimum length and the encounter time of minimum duration.

Now let's imagine what the moving man sees in his own inertial frame. He sees the bus, initially at distance ##b##, move towards him at constant speed in a straight line; the distance between him and the bus diminishes until it becomes zero.

The man's initial velocity (speed and direction) in the inertial frame of the ground, ##\mathbf v_m##, determine the encounter time. The initial separation ##b## in the frame of the man will diminish slower if ##\mathbf v_m## in the ground frame has a component along ##b##. It follows that if ##\mathbf v_m## is perpendicular to ##b##, the encounter time and hence the speed ##v_m## will be minimized.
 
  • #10
kuruman said:
I will rephrase what I tried to say because it didn't come out right. Consider the situation in which the man encounters the bus after selecting an appropriate velocity for that to happen. His path is a straight line angled as shown in the diagrams above. At minimum speed, his path will be of minimum length and the encounter time of minimum duration.

Now let's imagine what the moving man sees in his own inertial frame. He sees the bus, initially at distance ##b##, move towards him at constant speed in a straight line; the distance between him and the bus diminishes until it becomes zero.

The man's initial velocity (speed and direction) in the inertial frame of the ground, ##\mathbf v_m##, determine the encounter time. The initial separation ##b## in the frame of the man will diminish slower if ##\mathbf v_m## in the ground frame has a component along ##b##. It follows that if ##\mathbf v_m## is perpendicular to ##b##, the encounter time and hence the speed ##v_m## will be minimized.
Right, so it is perpendicular to the line joining the start points, not perpendicular to the highway.
 
  • #11
kuruman said:
At minimum speed, his path will be of minimum length and the encounter time of minimum duration.
Those three minima are not the same thing.

The minimum path length is for a path that goes directly toward the road. i.e. in a direction perpendicular to the road. In order to arrive at the road before the bus, this will require more speed than some other paths.

The minimum encounter time (minimum clock reading at the time of the intercept) is for a path that goes directly toward the bus in zero time. This requires infinite speed, so it is not an achievable minimum.

The minimum speed (while still arriving prior to the bus) is for a path that goes perpendicular to the line toward the bus.

As always with an optimization problem, three things need to be made clear. What value are we optimizing for, what constraints are being held fixed and what things are allowed to vary?
 
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  • #12
kuruman said:
At minimum speed, his path will be of minimum length and the encounter time of minimum duration.
Not true. The encounter time need not be of minimum duration, nor the path of minimum length. Only the (variable) velocity must be minimum. [If one is allowed to choose his velocity, but not the other].

Consider the problem in this thread. If the man walked perpendicular to the highway, his length is minimum, so is that for the bus. But will he travel at the least possible speed for meeting? No.

Consider the motion of the man along the path perpendicular to the initial line of separation b. Is it the least path he can take for a meeting? No. The least time? Also no. But the least speed? Yes.

I understand the strange nature of the problem, which is why we should think this over. After all, if times are fixed (for a meeting to happen at the same time), then shouldn't ##\Delta x=v\Delta t## imply that for minimum ##v## we should have minimum ##x##? Yes in one dimensions, but that's not the case here.
 
  • #13
brotherbobby said:
Incidentally, upon putting ##\color{blue}{(1)}## as a trigonometric equation in Mathway for it to solve for me, I got a different answer. It tells me the angle ##x=0.8^{\circ}, 2.64^{\circ}##, which are both far from my answer of ##\theta_1=45.6^{\circ}##.

Request : Where am I going wrong in evaluating the solution for the equation online?
It seems like Mathway isn't giving the answer in degrees but in radians. LLMs (large language models) aren't perfect and often won't give the exact response you're looking for, so use them with caution.
 
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  • #14
I suppose it is clear to those who have been following this thread that, despite its elementary nature, the problem I posed as the OP is not conceptually trivial. In order to clear some of the confusion, I have decided to re-do the problem with an eye on the questions of (1) least time, (2) least distance and (3) least velocity. The numbers (data) from the text were unwieldy to afford clear comparison, so I have simplified them.

A bus B travels at a speed of 4 m/s along the highway BC. Initially, a man is located at a distance 6 m from the highway and 10 m from the bus.
1715448613452.png


(I) Minimum speed : With what minimum speed (##v_m##) and in what direction (##\theta##) should the man walk in order to meet the bus along the highway? Also calculate the time and distance travelled by the man.

Since they both meet after the same time ##t##, ##\small{\dfrac{MC}{v_m}=\dfrac{BC}{v_B}\Rightarrow v_m=\dfrac{MC}{BC}v_B}##. But ##\small{\dfrac{MC}{BC}=\dfrac{\sin(\angle MBC=\theta)}{\sin\alpha}\Rightarrow v_m=\dfrac{\sin\theta}{\sin\alpha}v_B}##. For ##v_m## to be minimum, ##\sin\alpha## should be maximum, hence ##\boxed{\boldsymbol{\alpha=\pi/2}}##. If ##\small{\alpha=\pi/2\Rightarrow\angle AMC=\theta=\sin^{-1}0.6\approx 37^{\circ}}##. This is the direction of travel. The meeting point is at C, shown by the blue cross (##\color{blue}{\boldsymbol{\times}}##).
##\small{AC = 6\tan\theta=6\times 3/4=4.5\text{m}}##. Hence, the distance travelled by the bus, ##\small{BC = AB +AC = 8+4.5 = 12.5\text{m}}##. The time taken by the bus ##\small{t=\dfrac{12.5}{4}=\boxed{\boldsymbol{3.125\text{s}}}}##. This will also be the time taken by the man, in order that they both meet. The distance travelled by the man : ##x_M = \small{MC = 6\sec\theta=6\times\dfrac{5}{4}=\boxed{\boldsymbol{7.5\text{m}}}}##. Hence, the minimum speed by the man ##\small{\color{red}{v_m}=\dfrac{7.5}{3.125}=\boxed{\color{red}{\boldsymbol{2.4\text{m/s}}}}}^{\color{blue}{\large{1}}}##.


1715450245683.png
(II) Minimum distance : What is the minimum distance that the man should travel in order to meet the bus? Also calculate the speed of the man, the direction and the time he will take.

It is obvious from the diagram that the direction for minimum distance is the path perpendicular to the highway, which I have shown in brown ##\color{brown}{\text{dotted line}}##. The meeting point is at A, shown by the blue cross (##\color{blue}{\boldsymbol{\times}}##). The time taken by the man is the same as that for the bus : ##\small{t=\dfrac{8}{4}=\boxed{\boldsymbol{2\text{s}}}}##. The minimum distance that will be travelled by the man is ##\small{\color{red}{x_M = \boxed{\boldsymbol{6\text{m}}}}}##. The speed of the man ##\small{v_m = \dfrac{6}{2}=\boxed{\boldsymbol{3\text{m/s}}}}^{\color{blue}{\large{2}}}##.


1715451284576.png
(III) Minimum time : Now this is tricky but also the least interesting of the three cases. Minimum time with what speed? If the man is allowed to walk at any speed he chooses, then the best he can do is to travel at infinite speed along MB, which I have shown in brown ##\color{brown}{\text{dotted line}}##. The meeting point is at B, shown by the blue cross (##\color{blue}{\boldsymbol{\times}}##). Hence his speed ##\small{v_m=\boxed{\boldsymbol{\infty}}}## and his time is clearly ##\small{\color{red}{t=\boxed{\boldsymbol{0}}}}##. The distance travelled by the man ##\small{x_M = \boxed{\boldsymbol{10\text{m}}}}## and the angle at which he should walk is the angle with the highway that the line joining him to the bus made initially ##\small{\theta=\boxed{\boldsymbol{37^{\circ}}}}##.

1715448613452.png
Note :
The superscripts (1) and (2) which indicate the speeds for minimum speed and minimum distance. That the speed (1) is less than the speed (2) (for minimum distance) shows that the man should not move perpendicular to the highway if he wants to meet the bus with minimal speed. I hope this clarifies the doubt we have been having, at least numerically. Conceptually is another matter.
 
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  • #15
docnet said:
It seems like Mathway isn't giving the answer in degrees but in radians.
I see that, yet I had explicitly chosen "degrees" when Mathway asked me. It is good that it did, but failed to deliver on my request. 😥
 
  • #16
brotherbobby said:
(III) Minimum time : Now this is tricky but also the least interesting of the three cases. Minimum time with what speed? If the man is allowed to walk at any speed he chooses, then the best he can do is to travel at infinite speed along MB
If, instead, the man's speed is fixed but his walking angle is left variable and if we only require that he reaches a point on the road before the bus reaches that point then we have a simple intercept problem (aka a torpedo firing solution problem).

The man wants to reach the road just as the bus gets to the same point. Otherwise, he will waste time loitering on the roadside and could have done better by reducing his lead angle. Hence, the man must be shooting for an exact intercept.

If an intercept is possible there will (usually) be two possible intercept angles. Without doing the math, it feels like there is a quadratic in there. If the discriminant is negative, no intercept is possible. The minimum time is the time for the earlier of the two intercepts.
 
  • #17
jbriggs444 said:
If, instead, the man's speed is fixed but his walking angle is left variable and if we only require that he reaches a point on the road before the bus reaches that point then we have a simple intercept problem (aka a torpedo firing solution problem).

The man wants to reach the road just as the bus gets to the same point. Otherwise, he will waste time loitering on the roadside and could have done better by reducing his lead angle. Hence, the man must be shooting for an exact intercept.

If an intercept is possible there will (usually) be two possible intercept angles. (Without doing the math, it feels like there is a quadratic in there). The minimum time is the time for the earlier of the two intercepts.
Please refer to my post #1 (original problem). I have solved the matter you discuss.

More directly than there, if the man is given a speed to walk and he is asked to meet the bus at a minimum time, he has no choice of location or time. There is only a certain point along the highway where he will meet the bus exactly as it arrives. Before it, the bus arrives earlier - following it, the man arrives earlier.

Of course there is another point on the other side of "the divide" for a meet, but that would take a longer time for both of them.
 
  • #18
docnet said:
It seems like Mathway isn't giving the answer in degrees but in radians. LLMs (large language models) aren't perfect and often won't give the exact response you're looking for, so use them with caution.
Right.

##0.8## radians ##\approx 45.8^\circ## . Of course, the given value for the solution for ##x## has only one significant figure. It should be something closer to ##0.7767## radians.
 
  • #19
brotherbobby said:
I see that, yet I had explicitly chosen "degrees" when Mathway asked me. It is good that it did, but failed to deliver on my request. 😥
It's likely that the LLM inputed your prompt into a symbolic solver (such as Wolfram) that returned the answer in radians, and the answer was not converted into degrees by the LLM before returning it to the user (you). This probably happened because the LLM didn't "know" that the symbolic solver uses radians to compute the trig functions as a default.

The goal of LLMs in general is to understand the implicit and explicit meanings of user queries, and correctly predict the right answer. In your case the model 'hallucinated' (its a technical term for when a LLM gives the wrong answer) because you already made it clear in the prompt that you wanted the answer in degrees. And if you repeated the query and made it specific by stating "find ##x## in ##395-60\cot x=240\csc\ x##, and then apply the formula ##x^*= \frac{180}{\pi}x##, you may have gotten the right answer. LLMs that integrate symbolic solvers are trained on the behaviors of the symbolic solvers, and they are probably not that advanced yet. And so I believe it's better to directly use a symbolic solver, rather than an LLM that integrates one, to avoid the chance of the LLM making a mistake.
 
  • #20
brotherbobby said:
Now this is tricky but also the least interesting of the three cases.
If you find it less interesting it's because you made it so by departing from the original statement which says "The man can run at a speed of ##v_2=4\,\text{m/s}##." This is one-quarter of the speed of the bus and it is reasonable to assume that it is the man's top speed. It seems to me that you need an argument that takes into account a finite upper limit to the man's speed, preferably less than the speed of the bus.
 
  • #21
kuruman said:
If you find it less interesting it's because you made it so by departing from the original statement which says "The man can run at a speed of v2=4m/s."
That was the first part of the problem (see original post# 1). In the second part of the problem, the man's speed is not given. The speed has to be found given certain conditions. The authors gave minimum speed of man M to meet as the condition. I considered others like minimum time and minimum distance too.
 

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