What is the perceived relative velocity of these two motorcycles?

[Badmeditator]

Homework Statement

2 motorcycles accelerate uniformly from a common starting point to a semicircular finishing line 200 meters away. Their paths diverge, with an angle of 30° between them. They both cover the distance in 15 seconds. What is the velocity of motorcycle 1 from the perspective of motorcycle 2?

Relevant Equations

cos \theta = adj/hyp
v = d/t

I tried to project one on top of another but I know that's incorrect since I don't know how of do this.

 

[.Scott]

Follow these steps:
1) Calculate the acceleration needed to start from 0m/s and reach 200 meters in 15 seconds.
2) Using a Cartesian coordinate system with one axis running straight through the center line between the tracks of both motorcycles, compute the x/y velocities of each motorcycle.
3) The results will be a shared (same) velocity along one axis and opposite velocities along the other. You should know what to do from there.

 

[haruspex]

I tried to project one on top of another but I know that's incorrect since I don't know how of do this.

Not sure what you mean by projecting one on the other, but if you do not know how to do it, how do you know your approach is incorrect?
Can you explain more what you had in mind?

The question is somewhat unclear. They were both accelerating, so what relative velocity is being sought? Average? Final? Or does it ask for relative acceleration?

 

[jbriggs444]

I agree on the clarity. If I were handed this question, I would compute the velocity of each motorcycle relative to the other as a function of time.

A number of approaches are possible. Rather than setting up a coordinate system where both motorcycles move at angles, my inclination would be to set up a coordinate system where one motorcycle moves vertically along the y-axis and the other does whatever it does as measured against those coordinates.

Then one could repeat for the other motorcycle or invoke a simple argument from symmetry.

The problem statement does not provide guidance on how the result is to be reported. However, since we are asked about "from the perspective of", my inclination would be to report magnitude and direction with direction measured relative to the observing motorcycle's heading.

Oh, of course...

Make a drawing and, if possible, share it with us. With a good drawing in hand, many problems become simple.

 

[.Scott]

There is another approach.
1) Calculate the motorcycle positions at t=15 seconds and subtract cycle 1 position from cycle 2 position.
2) Calculate the distance between the motorcycles at t=15 seconds. I find this most easily done without reference to the results from step 1.
3) Calculate the required constant acceleration to reach that distance in 15 seconds.
4) Calculate the unit direction vector from cycle 1 to cycle 2 by dividing the result of 1) by the result of 2).
5) Multiply 3 and 4 to give you a constant acceleration vector.

 

[Badmeditator]

Not sure what you mean by projecting one on the other, but if you do not know how to do it, how do you know your approach is incorrect?
Can you explain more what you had in mind?

The question is somewhat unclear. They were both accelerating, so what relative velocity is being sought? Average? Final? Or does it ask for relative acceleration?

The question itself was very vague. This was on a test and I had no clue how to proceed so I just tried to find a vector of a motorcycle and project it by multiplying it with cos 30. Obviously it came out wrong.

 

[jbriggs444]

You could indeed look at the vector of the other motorcycle projected onto the path of the one. That would involve multiplying by $\cos 30$ just as you did. This would have given you the forward component of the other motorcycle's velocity in the ground frame.

But you were not asked for a velocity in the ground frame. You were asked for a relative velocity. So you have to subtract off the forward velocity of the motorcycle you are on.

That would give you the forward component of the other motorcycle's velocity relative to the one. What about the sideways component?

You can do the same analysis for the sideways direction, multiplying the other motorcycle's velocity by $\sin 30$ to get its sideways velocity in the ground frame.

That would be a ground-relative velocity. But subtracting your own motorcycle's sideways velocity (zero) would not change it at all.

 

[kuruman]

By symmetry, as the motorcycles are headed to the endpoints of the semicircle, the components of the two velocities along the line from the starting point to the center of the semicircle are equal at all times. The correct answer would be phrased as "the relative velocity is $v$ m/s away from the observer motorcycle along the diameter of the semicircle."

The SUVAT equation to use is $\Delta s=\frac{1}{2}(v_0+v_f)\Delta t$ and the rest is trigonometry.

 

[PeroK]

Homework Statement:: 2 motorcycles accelerate uniformly from a common starting point to a semicircular finishing line 200 meters away. Their paths diverge, with an angle of 30° between them. They both cover the distance in 15 seconds. What is the velocity of motorcycle 1 from the perspective of motorcycle 2?
Relevant Equations:: cos \theta = adj/hyp
v = d/t

I tried to project one on top of another but I know that's incorrect since I don't know how of do this.

I would take the x-axis to split the angle between the paths. That means one motorcycle is accelerating on a path at $+15$ degrees and the other at $-15$ degrees.

By symmetry, the velocity components in the x-direction are the same. The relative velocity, therefore, is the relative velocity in the y-direction and has a magnitude of twice the y-component of either motorcycle.

 

[.Scott]

Let's say that the motorcycles start at the origin of the coordinate system $(0,0)$; the bearing of Motorcycle 1 is as $\beta$; and the bearing of Motorcycle 2 is $(\beta+30^{\circ})$.

The bearing of Motorcycle 2 relative to Motorcycle 1 can be probably be determined by thought - or perhaps with the assistance of a sketch.

Then, all that is needed is the constant rate of apparent acceleration. Since we already know the travel time (15 seconds), all that is needed is the rate of acceleration - which can be determined by the distance between the motorcycles at t=15. There is no cosine or $30^{\circ}$ in the solution.

 

[kuruman]

Let's say that the motorcycles start at the origin of the coordinate system $(0,0)$; the bearing of Motorcycle 1 is as $\beta$; and the bearing of Motorcycle 2 is $(\beta+30^{\circ})$.

The bearing of Motorcycle 2 relative to Motorcycle 1 can be probably be determined by thought - or perhaps with the assistance of a sketch.

Then, all that is needed is the constant rate of apparent acceleration. Since we already know the travel time (15 seconds), all that is needed is the rate of acceleration - which can be determined by the distance between the motorcycles at t=15. There is no cosine or $30^{\circ}$ in the solution.

You don't need the acceleration. Please see posts #8 and #9. Also, it would be easier to consider the origin where you have it and let the y-axis be the line joining the origin with the center of the semicircle. Then the bearings would be 15^o on either side of the y-axis. Each motorcycle sees the other move along the x-axis away from it. Yes, a sketch is important.

I haven't seen the solution, but if there should be no $\cos30^o$ in it. If you think there ought to be, it's because you have not properly analyzed the problem. We have given you all the hints you need. Try to understand what they are saying to you.

 

[jbriggs444]

You don't need the acceleration. Please see posts #8 and #9. Also, it would be easier to consider the origin where you have it and let the y-axis be the line joining the origin with the center of the semicircle. Then the bearings would be 15^o on either side of the y-axis. Each motorcycle sees the other move along the x-axis away from it. Yes, a sketch is important.

I haven't seen the solution, but if there should be no $\cos30^o$ in it. If you think there ought to be, it's because you have not properly analyzed the problem. We have given you all the hints you need. Try to understand what they are saying to you.

There can be a cos 30 in it if you choose a coordinate system where a chosen motorcycle moves along a coordinate axis. Depending on how you choose to report the result, that cos 30 may or may not disappear.

Suppose, for instance, you choose to report the result as an $(v_x,v_y)$ coordinate pair where the y-axis is lined up on the local motorcycle's path and the other motorcycle is on a path 30 degrees to the right of the y axis.

Suppose that we have computed the magnitude of the two motorcycle's ground-relative velocities. $\vec{v}$. [Possibly we have used the SUVAT equations to obtain $|\vec{v}|$ as a simple function of time].

We can immediately see that the ground-relative velocity of the remote motorcycle is given by $(v \sin 30, v \cos 30)$. Even more easily we see that the ground-relative velocity of the local motorcycle is given by $(0,v)$.

The local-motorcycle-relative velocity of the remote motorcycle is given by the difference between these two ground-relative velocities: $(v (1 - \cos 30), v \sin 30)$.

Now, of course the magnitude of this result must match the magnitude of the result calculated using the coordinate system whose y-axis splits the difference between the two motorcycle paths.

That is, the length of the chord of a 30 degree circular arc of a unit circle will be the same whether computed as $2\sin 15$ or as $\sqrt{(1-\cos 30)^2 + \sin^2 30}$. That result is 0.517 either way.

 

[kuruman]

There can be a cos 30 in it if you choose a coordinate system where a chosen motorcycle moves along a coordinate axis. Depending on how you choose to report the result, that cos 30 may or may not disappear.
$\dots$

Yes, of course. OP has considered an even more general case than that in which the ground velocities would be $\{v\sin\beta,v\cos\beta\}$ for the local motorcycle and $\{v\sin(\beta+30^o),v\cos(\beta+30^o)\}.$ In that context, any specific angle $\beta$ can appear in the equation, but that would complicate things unnecessarily.