Acceleration: 3 people walk toward each other

[Raghav Agarwal]

Homework Statement

Three particles A, B, C, are situated at the vertices of equilateral triangle of side l. They approach each other with velocity v, such that A always moves toward B, B towards C and C towards A. What will be the initial acceleration?

Homework Equations

The Attempt at a Solution

Using vector components, I have found the acceleration
a=v_approach*(rate of change of direction of velocoty vector
a=(3v/2)*(dθ/dt)

However I could not calculate this rate. I have tried to form trigonometric relations and differentiate them but it failed, and I am out of any new ideas.

 

[kuruman]

Are we to assume that the speed v of the people (or is it particles?) is constant throughout the motion? If so, what kind of acceleration, other than zero, maintains constant speed?

 

[haruspex]

a=V_approach*(rate of change of direction of velocoty vector

Why the relative velocity ?

However I could not calculate this rate.

Imagine yourself as A, headed towards B. Does the fact that B is not moving in exactly a straight line make much difference to you initially? So how fast will you be changing direction?

 

[DaveE]

Are you sure that you have stated the problem correctly? As I read it: A starts with velocity v towards B and is constrained to travel towards B. The question "what is the initial acceleration on A?" seems trivial to me. I suspect that I am misinterpreting something.

 

[kuruman]

Are you sure that you have stated the problem correctly? As I read it: A starts with velocity v towards B and is constrained to travel towards B. The question "what is the initial acceleration on A?" seems trivial to me. I suspect that I am misinterpreting something.

If by "trivial" you mean "zero" as in trivial solution, that is not the case. The entities (people or particles) change their velocity (assumed instantaneously) so that it is always directed towards the entity in front of them.

 

[DaveE]

If by "trivial" you mean "zero" as in trivial solution, that is not the case. The entities (people or particles) change their velocity (assumed instantaneously) so that it is always directed towards the entity in front of them.

Yes, I meant zero. But now I see that you need initial acceleration to avoid a pointing error in the next instant. Doesn't seem causal, but no one said it had to be, I think that was what confused me.

 

[Raghav Agarwal]

Are we to assume that the speed v of the people (or is it particles?) is constant throughout the motion? If so, what kind of acceleration, other than zero, maintains constant speed?

Yes the magnitude of their velocity remains constant but the direction changes, like in uniform circular motion

 

[Raghav Agarwal]

Why the relative velocity ?

Imagine yourself as A, headed towards B. Does the fact that B is not moving in exactly a straight line make much difference to you initially? So how fast will you be changing direction?

I have tried assuming that they move dl distance along the stright line in dt time, ignoring the direction change. Then I used co-ordinate geometry to find the new slope and calculate d(theta). I was left with a value that had terms of dl and dt, and hence useless

 

[Raghav Agarwal]

I made another attempt at it.
I assumed that the triangle that forms between the three, rotates with a constant angular velocity. This too yielded nothing.
This is the working for that https://m.imgur.com/gallery/9X34cs1

 

[haruspex]

I have tried assuming that they move dl distance along the stright line in dt time, ignoring the direction change. Then I used co-ordinate geometry to find the new slope and calculate d(theta). I was left with a value that had terms of dl and dt, and hence useless

Please post your working for that attempt.
And please do it by typing in the post. Images are often hard to read, and the link in post #9 doesn’t work for me.

 

[kuruman]

Yes the magnitude of their velocity remains constant but the direction changes, like in uniform circular motion

Precisely. So what is the acceleration for uniform circular motion? Draw a picture.

 

[Raghav Agarwal]

Please post your working for that attempt.
And please do it by typing in the post. Images are often hard to read, and the link in post #9 doesn’t work for me.

I assume that for dt time they don't change their direction and move in straight line.
$$\text{Initial slope}=\sqrt{3}$$
$$\text{Slope after }dt=m _2$$
$$d\theta= \tan^{-1}
\left(\frac
{l*\frac sqrt{3} 2 -dl \sqrt{3}} {\frac l 2}}-sqrt{3}\right)$$

Precisely. So what is the acceleration for uniform circular motion? Draw a picture.

It is V^2/r, but i will need radius of curvature to do that here, for which i will require the eqn of their path, which i haven't got. I tried doing it in reverse assuming angular velocity constant but that led nowhere.
Do you have any idea for the path followed

 

[kuruman]

Draw a picture. The entities form an equilateral triangle at all times and are moving with speed v at all times. You are interested in the particular time when the side of the triangle is ##l##. Draw a picture.

 

[Raghav Agarwal]

Draw a picture. The entities form an equilateral triangle at all times and are moving with speed v at all times. You are interested in the particular time when the side of the triangle is ##l##. Draw a picture.

I have drawn it multiple times, I am not trying to solve the problem in my head. Even the book has a detailed picture
Here is one I found on the web, just to show that i know what you are saying
http://2.bp.blogspot.com/-lLb0RcwG_LQ/UKpPDbIYQ-I/AAAAAAAAB3s/Yklf-9iXLQ8/s1600/scan00014.jpg
Please help with something a little less cryptic.

 

[Raghav Agarwal]

Here is the working so far

 

[kuruman]

Please do this.
Draw a nice big equilateral triangle and label its side ##l##. Choice one of the three entities and draw an arrow representing its velocity vector and label it ##\vec v.## Also draw its acceleration vector and label it ##\vec a##. Be sure that you draw the acceleration vector correctly. Correctly means that it changes the direction of the velocity but not the magnitude (speed). What could the magnitude of ##\vec a## be?

 

[Raghav Agarwal]

So a will be really small, making an angle slightly less than 90, to not change the magnitude
Exact angle would require the angle between ## \vec V ## and ##\vec V'##

Of course this is for vector addition of V and a*t, but i could draw no other conclusions from this

 

[haruspex]

I assume that for dt time they don't change their direction and move in straight line.
$$\text{Initial slope}=\sqrt{3}$$
$$\text{Slope after }dt=m _2$$
$$d\theta= \tan^{-1}
\left(\frac
{l*\frac sqrt{3} 2 -dl \sqrt{3}} {\frac l 2}}-sqrt{3}\right)$$

You have made a mistake in the latex and I cannot figure out what you meant.
But I think there is an easier way than working with slopes.
Suppose B has moved to position B'. AB and AB' will be almost the same length. (You can make an approximation for for the difference if you like.) using the lengths AB' and BB' and the angle ABC, what expression can you write for angle BAB'?

 

[Raghav Agarwal]

You have made a mistake in the latex and I cannot figure out what you meant.
But I think there is an easier way than working with slopes.
Suppose B has moved to position B'. AB and AB' will be almost the same length. (You can make an approximation for for the difference if you like.) using the lengths AB' and BB' and the angle ABC, what expression can you write for angle BAB'?

I have tried fixing that latex mistake But i can't find it. I have written it out and posted in #15, this time a bit more clearer.

Using that
$$d\theta=vdt/l$$
$$d\theta/dt=v/l$$

Thanks, I will verify the answer tommorow from our teacher

 

[haruspex]

I have tried fixing that latex mistake But i can't find it. I have written it out and posted in #15, this time a bit more clearer.

Using that
$$d\theta=vdt/l$$
$$d\theta/dt=v/l$$

Thanks, I will verify the answer tommorow from our teacher

I was thinking of the sine rule: vdt/sin(dθ)=AB'/sin(60)≈AB/sin(60).
And sin(dθ)≈dθ.

 

[ehild]

Because of symmetry, the three points are always on the corners of concentric equilateral triangles, shrinking and turning in time. I would work with polar coordinates.

The shaded angle is 30°, so the radial component of the velocities is ##\dot r= v\cos(30)##, the angular component is ##r\dot \theta = v\sin(30)##. Use the expression of the acceleration in polar coordinates. https://en.wikipedia.org/wiki/Polar_coordinate_system#Vector_calculus

 

[Raghav Agarwal]

I was thinking of the sine rule: vdt/sin(dθ)=AB'/sin(60)≈AB/sin(60).
And sin(dθ)≈dθ.

Thanks for the confusion.
Seriously, which ans is more accurate?
I used $$\theta=arc/radius$$

 

[Raghav Agarwal]

I am not familiar with calculus in polar coordinates but i want to try.
What does that dot on top mean, is it first degree diffrentiation, like f'(x)?

 

[ehild]

I am not familiar with calculus in polar coordinates but i want to try.
What does that dot on top mean, is it first degree differentiation, like f'(x)?

Yes, the dot on the top means differentiation with respect to time. ##\dot r = \frac {dr}{dt}## and ##\dot \theta=\frac {d\theta}{dt}##.

 

[haruspex]

Thanks for the confusion.
Seriously, which ans is more accurate?
I used $$\theta=arc/radius$$

What confusion?
In post #20 I have given you equations by which you can very quickly find dθ/dt.