# Homework H1.I - Sp23

 Problem statement Solution video

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DISCUSSION
This is a standard relative motion type of question. You are given complete information on the velocity vectors for A and P (both x- and y-components). And, you are asked to find the velocity of P with respect to A.

To do so, use the basic equation for relative velocity between two points:  vP/A = vP - vA. From this result, you are then asked to find the angle of the observed velocity of P, with this last step being just trig.

## 28 thoughts on “Homework H1.I - Sp23”

Is (b) asking for the angle that the passenger observes P at? Not entirely sure what “with the vertical” means in regards to the angle.

1. Benjamin Carl Wassgren says:

The vertical is straight along the y axis. It shouldn't change when going from the stationary to moving point of view (the thing that changes is the direction and magnitude of the velocity vector).

2. CMK says:

In the figure, you see that the velocity of P makes an angle theta_P with the vertical. Same for the angle you are asked to find: the angle that v_P/A makes with the vertical.

Would this mean you are taking the v_P/A equation and plugging in a theta value in v_P/A that you are solving for? I don't quite understand the process with this.

1. CMK says:

I am not quite sure of your question. Find the vector v_P/A = v_P - v_A. Using the x- and y-components of this equation and a little trig, you can find the angle theta_P/A.

2. Victoria Taylor Knott says:

My thought process for B was similar to how I approached A, with thetaP/A being equal to thetaP - theta A ; did anybody else think through it this way?

1. Jackson Charles Miller says:

Wouldn't that mean the angle observed was the same as theta_P. Does that make sense?

That thought process makes the most sense to me. You are seeing the angle of P from the automobile A.

2. Quan Minh Huynh says:

I don't think that thetaP/A = thetaP - thetaA because in my understanding, this would only work for vectors.

3. Elijah Allan Collins says:

I was thinking the angle would just be 90? Is that the right thinking? It felt to simple and maybe I am understanding the problem wrong.

1. Elijah Allan Collins says:

or would it just be 90-theta_p, I am not sure and would appreciate any constructive feedback on how others approached B.

1. Jenna says:

Since it is asking for the angle with the vertical, I just drew out the vector I got from part A. Then I used trig and the vector components (inverse tangent) to find out what angle it made with the vertical.

2. Anna Grace LiHua Hook says:

Since the question is asking for relative velocity the angle in the picture is going to be different than the angle as viewed by the observer. Once you use the relative motion equation you have new i and j components making up the relative velocity vector. By trig you can then see what angle the resultant vector makes with the vertical.

4. Phillip Eugene Holmes says:

My method for part b was to use the equation Vp/a bar = Vp bar - Va bar. From here i could equate the vectors to their magnitude multiplied by the angle of direction for the i and j components. knowing this and the value for Vp/a from part a i reduced the values down and solved for theta. This makes sense to me is there anything wrong with this thought process?

5. Pause says:

For part a, do you use the given theta value to get scalar values for your velocity of P with respect to A? Because my thought process was from there you can use the i and j components of this vector to solve for the other theta value in part b.

Yeah that’s what I ended up doing. Converted the given velocities to Cartesian components, then used Vp/a = Vp - Va from there.

6. Blake Charles Boyer says:

If I calculated the angle using the cos(x) identity between both vectors do I need to subtract the angle I found from 90? Subtracting the angle I found from 90 seems like the only way to get the angle with respect to the vertical axis.

1. Alicja Stoppel says:

I didn't use the cosine identity, I used arctan to find it, but in the end I also had to subtract 90 to get it with the vertical so I believe you probably should do the same.

7. Trysta Chiang says:

For part a, could as assume Vp/a's cartesian components have both i and j components?

1. CMK says:

It is not really an assumption; it follows from writing down the vector equation: v_P/A = v_P - v_A. You naturally end up with both x- and y-components.

8. Trisha Boodhoo says:

I have heard the inverse tangent being mentioned a lot for part (b). I used the velocity vector and divided it into its i and j components such that i: v(p/a) = |v(p/a)|cos(theta) and j: v(p/a) = |v(p/a)|sin(theta). Then I find the key angle in the i component using inverse cosine, which gives me the same answer as the key angle from the j component using inverse sine. I don't understand where an inverse tangent comes into play?

1. CMK says:

Any one of those three approaches will work: inverse sine, inverse cosine or inverse tangent.

For inverse sine, you would use the hypotenuse and the x-component.

For the inverse cosine, you would use the hypothenuse and the y-component.

For the inverse tangent, you would use both the x- and y-components.

It's just trig.

1. Tate Peters says:

I found it easiest to use the inverse tangent as we have already solved for the x and y components. Just do tan-1(i/j) to solve for theta.

9. Anthony Michael Cialdella says:

When trying to find the angle in part B, can we assume that the car has a velocity of 0, and that the entire xcomponent of velocity found in part A is solely due to P?

1. CMK says:

No, that would not be a very good plan. Simply use the vector equation: v_P/A = v_P - v_A. Then, find the angle from the x- and y-components of that result.

2. Lila Johnson says:

I'm not sure if this is entirely correct but this is what I did: I drew v_P/A found in part a) as a vector next to a vertical line to make an angle, theta. I then used the x and y (or i and j) components of v_P/A to make a triangle. The x-component was opposite of the angle while the y-component became the adjacent side, and the v_P/A vector became the hypotenuse. From there I used those lengths of the triangle and trig to find the angle.

10. Peter Thomas Hays says:

When I solved this problem the first time, I used cosine for the horizontal component and sine for the vertical as usual. But the diagram has it flipped so vertical will be cosine and vice versa, hopefully this doesn't trip anyone else up.