- #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.##

**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.

(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}}}}##.

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).

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?*

(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}}}##.

*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##.

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?*

**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}}}##.

*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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