I Road crossing problem from AEH Love's text

[chms]

TL;DR Summary

avoiding collisions whilst crossing road and stream of busses

AEH Love is well known for his Treatise on Elasticity. He has also authored a text titled "Theoretical Mechanics An Introductory Treatise on the Principles of Dynamics"; published in 1906. Problem sets in it are challenging and I have been puzzling over this one from the first set of "Miscellaneous Examples":

Prove that the time in which it is possible to cross a road of breadth $c$ in a straight line, with the least uniform velocity, between streams of busses of breadth $b$, following at intervals , moving with velocity $V$, is $\frac{V}{c}(\frac{a}{b} + \frac{b}{a})$

Each bus is length $b$ , and if we start from the edge of the road aligned with the front of the bus, we need to traverse a diagonal distance $\sqrt{(a+b)^2 +(c/2)^2}$ in time $(a+b)/V$
Which will bring us to the midline and aligned with the tail end of the first gap between the bus whose front we were aligned with and the bus behind it.
This gives a velocity of $\frac{V}{a+b}\sqrt{(a+b)^2 +(c/2)^2}$ which is also used to traverse the remaining distance of
$\sqrt{(a+b)^2 +(c/2)^2}$
This gives a total time of $2(a+b)/V$, which is different from what we have been asked to show. What I am I missing?

 

[mfb]

What is a? Why did you calculate something until the middle of the road crossing instead of the full road crossing?

and if we start from the edge of the road aligned with the front of the bus

Wouldn't you start right after the bus passed you, i.e. at the end of the bus?

 

[A.T.]

The parameter b is called breadth and length of a bus. Is there a diagram?

 

[chms]

TL;DR Summary: avoiding collisions whilst crossing road and stream of busses

AEH Love is well known for his Treatise on Elasticity. He has also authored a text titled "Theoretical Mechanics An Introductory Treatise on the Principles of Dynamics"; published in 1906. Problem sets in it are challenging and I have been puzzling over this one from the first set of "Miscellaneous Examples":

Prove that the time in which it is possible to cross a road of breadth $c$ in a straight line, with the least uniform velocity, between streams of busses of breadth $b$, following at intervals $a$, moving with velocity $V$, is $\frac{V}{c}(\frac{a}{b} + \frac{b}{a})$

Each bus is length $b$ , and if we start from the edge of the road aligned with the front of the bus, we need to traverse a diagonal distance $\sqrt{(a+b)^2 +(c/2)^2}$ in time $(a+b)/V$
Which will bring us to the midline and aligned with the tail end of the first gap between the bus whose front we were aligned with and the bus behind it.
This gives a velocity of $\frac{V}{a+b}\sqrt{(a+b)^2 +(c/2)^2}$ which is also used to traverse the remaining distance of
$\sqrt{(a+b)^2 +(c/2)^2}$
This gives a total time of $2(a+b)/V$, which is different from what we have been asked to show. What I am I missing?

 

[chms]

What is a? Why did you calculate something until the middle of the road crossing instead of the full road crossing?Wouldn't you start right after the bus passed you, i.e. at the end of the bus?

Apologies, a is the gap between the busses (post edited). the problem asks for the least velocity, so we would start as early as possible

 

[chms]

The parameter b is called breadth and length of a bus. Is there a diagram?

There is no diagram. I interpreted b as the length of the bus.

 

[chms]

 

[Lnewqban]

Welcome!

Should interval a be interpreted as the distance between the front end of each bus to the front or to the rear end of the next one?

 

[A.T.]

And what is 'least uniform velocity'?

 

[Ibix]

And what is 'least uniform velocity'?

I think that just means the lowest velocityyou can have so you don't get clipped by the bus behind. You could go faster.

 

[chms]

Welcome!

Should interval a be interpreted as the distance between the front end of each bus to the front or to the rear end of the next one?

View attachment 360852

With that interpretation, the gap is $(a-b)$ and the total time to cross the road is $2a/V$ using the same arguments as before.

 

[marcusl]

If the breadth of the road is c, normal to its length, then the breadth b of a bus must also be measured normal to its direction of travel. (Thus b is bus width, not length.)

 

[TSny]

Consider the buses' frame of reference. It seemed much easier for me in this frame.

 

[bob012345]

I find it interesting that no one seems to be able to do this. I wonder if a student in the U.K. a century ago would have had such difficulty interpreting this problem?

 

[TSny]

Spoiler: Here is my solution.

Let the buses move East and assume the runner starts from some point on the south side of the road.

After some thought, it should be clear that for the runner to get through the gap at minimum speed, she will almost strike the leading bus at point $A$ and the trailing bus at point $B$.

$V$ is the speed of the buses. Let $w$ be the speed of the runner relative to the road and let $u$ be the speed of the runner relative to the buses.

$w$, $u$, and $V$ are related by the relative velocity triangle:

In the reference frame of the buses, the trajectory of the runner must be along the brown line:

Thus, the direction of the runner’s velocity in the bus frame is fixed by the angle $\theta$, where $\tan \theta = \dfrac b a$.

Let $\beta$ be the angle between the runner’s velocity in the bus frame and the runner’s velocity in the Earth frame. So, in the velocity triangle, we have

From the law of sines, $w = V \dfrac {\sin \theta}{\sin \beta}$. Since $\theta$ is fixed, the minimum value of $w$ occurs for $\sin \beta = 1$ or $\beta = \dfrac {\pi} 2$. That is, vector $\vec w$ is perpendicular to $\vec u$ when the runner barely makes it through the gap with minimum speed $w$.

With $\beta = \dfrac {\pi} 2$, $w = V \sin \theta$. The northern component of the runner’s velocity relative to the ground is $w \cos \theta = V \sin \theta \cos \theta$.

Thus, the time for the runner to cross the road is $\dfrac {c} {V \sin \theta \cos \theta}$.



From $\tan \theta = \dfrac b a$, we have $\sin \theta = \dfrac {b}{\sqrt{a^2+b^2}}$ and $\cos \theta = \dfrac {a}{\sqrt{a^2+b^2}}$. Therefore, the time is $\dfrac c V \dfrac {a^2+b^2}{ab} = \dfrac c V \left( \dfrac a b + \dfrac b a \right)$.

 

[chms]

Wow! many thanks

 

[chms]

I find it interesting that no one seems to be able to do this. I wonder if a student in the U.K. a century ago would have had such difficulty interpreting this problem?

I was wondering that myself!

 

[chms]

Spoiler: Here is my solution.

Let the buses move East and assume the runner starts from some point on the south side of the road.

View attachment 360996

After some thought, it should be clear that for the runner to get through the gap at minimum speed, she will almost strike the leading bus at point $A$ and the trailing bus at point $B$.

View attachment 360997

$V$ is the speed of the buses. Let $w$ be the speed of the runner relative to the road and let $u$ be the speed of the runner relative to the buses.

$w$, $u$, and $V$ are related by the relative velocity triangle:

View attachment 360998

In the reference frame of the buses, the trajectory of the runner must be along the brown line:

View attachment 360999

Thus, the direction of the runner’s velocity in the bus frame is fixed by the angle $\theta$, where $\tan \theta = \dfrac b a$.

Let $\beta$ be the angle between the runner’s velocity in the bus frame and the runner’s velocity in the Earth frame. So, in the velocity triangle, we have

View attachment 361000


From the law of sines, $w = V \dfrac {\sin \theta}{\sin \beta}$. Since $\theta$ is fixed, the minimum value of $w$ occurs for $\sin \beta = 1$ or $\beta = \dfrac {\pi} 2$. That is, vector $\vec w$ is perpendicular to $\vec u$ when the runner barely makes it through the gap with minimum speed $w$.

With $\beta = \dfrac {\pi} 2$, $w = V \sin \theta$. The northern component of the runner’s velocity relative to the ground is $w \cos \theta = V \sin \theta \cos \theta$.

Thus, the time for the runner to cross the road is $\dfrac {c} {V \sin \theta \cos \theta}$.



From $\tan \theta = \dfrac b a$, we have $\sin \theta = \dfrac {b}{\sqrt{a^2+b^2}}$ and $\cos \theta = \dfrac {a}{\sqrt{a^2+b^2}}$. Therefore, the time is $\dfrac c V \dfrac {a^2+b^2}{ab} = \dfrac c V \left( \dfrac a b + \dfrac b a \right)$.

Excellent!

 

[PeroK]

View attachment 360844

Here's an alternative solution, using some calculus:

Spoiler

If we take the bus's motion as the positive x-direction.

Relative to the bus, let the components of the runner's velocity are $(-v_x, v_y)$, where the criterion for just making it through between the buses is:$$\frac{v_x}{a} = \frac{v_y}{b}$$Relative to the road, the components of the runner's velocity are $(u_x, u_y)$. To minimise the overall speed, $u_x$ must be in the same direction as $V$, hence:
$$u_x = V - v_x, u_y = v_y$$The runner's velocity squared is:
$$u^2 = u_x^2 + u_y^2 = (V-v_x)^2 + \frac{b^2}{a^2}v_x^2$$We can minimise this by setting the derivative of $u^2$ with respect to $v_x$ equal to zero:
$$-2(V-v_x) + 2\frac{b^2}{a^2}v_x = 0$$Leading to:
$$v_x = \frac{a^2}{a^2 + b^2}V$$This gives:
$$u_y = \frac b a {v_x} = \frac{ab}{a^2 + b^2}V$$And, finally, the time required to cross the road is:
$$T = \frac c{u_y} = \frac c V \bigg(\frac {a^2 + b^2}{ab} \bigg ) = \frac c V \bigg(\frac a b + \frac b a \bigg )$$

 

[chms]