Why Is Calculating the Relative Angle Between Two Moving Cars Tricky?

[Gold3nlily]

Two highways intersect as shown in Fig. 4-46. At the instant shown, a police car P is distance dP = 700 m from the intersection and moving at speed vP = 70 km/h. Motorist M is distance dM = 510 m from the intersection and moving at speed vM = 51 km/h. What are the (a)x-component and (b)y-component of the velocity (both in km/h) of the motorist with respect to the police car? (c) For the instant shown in Fig. 4-46, what is the angle between the velocity found in (a) and (b) and the line of sight between the two cars?

(I can't put up picture but the intersection is on the origin. The police car is driving in the -x direction along the x axis. The other car is driving down the y axis.)

Vx=Vcos(/)
Vy=Vsin(/)
V-> = (Vx)i + (Vy)j
V = (Vx^2 = vy^2)^1/2
(/) = tan^-1 (Vy/Vx)

Okay, so I got the first part of the problem right.
Vx=70km/hr
Vy=-51 km/h
I get that these are the components b/c the cars are driving along the axes. But at first I thought that both components would be negative. Why is the x component positive when the police car is driving in the -x direction? Why is the y component negative if the x component is positive?

the last part gets weird... I have both of the components and I am looking for the angle. I plug them into the equation:
(/) = tan^-1 (-51/70)
and I get -36.1 degrees.
I realize this negative angle is not in the right quadrant. But if I wanted the angle between the x-axis and the line of sight between the two cars wouldn't that just be positive 36.1 b/c it would be a complementary angle to the one I found with the equation? This was wrong. I also tried adding 180 degrees in case it wanted the other angle with respect to the positive x-axis but 144 was also not right. How do I find the angle?edit:
I'm getting views but no replies. Did I ask my question wrong? Is it too long? If I did something wrong, can someone please let me know?

 

[lanedance]

Two highways intersect as shown in Fig. 4-46. At the instant shown, a police car P is distance dP = 700 m from the intersection and moving at speed vP = 70 km/h. Motorist M is distance dM = 510 m from the intersection and moving at speed vM = 51 km/h. What are the (a)x-component and (b)y-component of the velocity (both in km/h) of the motorist with respect to the police car? (c) For the instant shown in Fig. 4-46, what is the angle between the velocity found in (a) and (b) and the line of sight between the two cars?

(I can't put up picture but the intersection is on the origin. The police car is driving in the -x direction along the x axis. The other car is driving down the y axis.)

bit hard to say exactly without pic, but i would say based on the info below, both the cars are moving in the -ve axis direction (left and down)

however the question asks for the velocity of the motorist with respect to the police car, as the police car is moving left, from the police car the motorist appears to be moving right (+ve x dir'n). the motion in the y dir'n is still -ve.

 

[lanedance]

for the angle, it asks for the angle against the "line of sight", not against the axis

 

[Gold3nlily]

"both the cars are moving in the -ve axis direction (left and down)"
Yes. you're right, left and down.

Here is the picture, I hope this link works:
http://gold3nlily.shutterfly.com/pictures/11

Okay, so becasue this Vx is measured from the police cars perspective, Vx is now positive even though it is moving in the negative x direction? Would it help if I always mentally flip the axes in my mind when trying to figure out the signs. When I do that the police car is hypothetically moving in the positive y direction and the other car is hypothetically moving in the negative x direction (of course I would have to flip to axes back for the rest of the problem, but will this always work?).

 

[Gold3nlily]

Here is another picture involving the angles:
http://gold3nlily.shutterfly.com/pictures/14

The pink angle should be the one I'm looking for right? It is the one against the "line of sight" right?

The green angle is what I think the equation gave me: -36.1

The blue angle is 180- 36.1 = 144 (3 sig figs...)

That must mean that the pink angle should be positive 36.1.
Why is this wrong??

I entered all three of these angles and they were all wrong. Please help. Am I supposed to be looking for an angle not indicated in my drawing? maybe the angle between the other car and the y axis?

Thank you.

 

[lanedance]

"both the cars are moving in the -ve axis direction (left and down)"
Yes. you're right, left and down.

Here is the picture, I hope this link works:
http://gold3nlily.shutterfly.com/pictures/11

Okay, so becasue this Vx is measured from the police cars perspective, Vx is now positive even though it is moving in the negative x direction? Would it help if I always mentally flip the axes in my mind when trying to figure out the signs. When I do that the police car is hypothetically moving in the positive y direction and the other car is hypothetically moving in the negative x direction (of course I would have to flip to axes back for the rest of the problem, but will this always work?).

don't flip anything, just picture what the policeman would see, as teh policeman is moving left, it appears fropm his point of view that the motorist is moving right (plus the original down velcoity as well)

 

[lanedance]

Here is another picture involving the angles:
http://gold3nlily.shutterfly.com/pictures/14

The pink angle should be the one I'm looking for right? It is the one against the "line of sight" right?

The green angle is what I think the equation gave me: -36.1

The blue angle is 180- 36.1 = 144 (3 sig figs...)

That must mean that the pink angle should be positive 36.1.
Why is this wrong??

I entered all three of these angles and they were all wrong. Please help. Am I supposed to be looking for an angle not indicated in my drawing? maybe the angle between the other car and the y axis?

Thank you.

you need to find teh angle between "the line of sight" and "the relative velcoity". The pink angle looks only like the line of sight.

 

[Gold3nlily]

"it appears from his point of view that the motorist is moving right (plus the original down velocity as well")

OMG! I get it- moving right AND down! Okay, yay!

 

[Gold3nlily]

you need to find the angle between "the line of sight" and "the relative velcoity". The pink angle looks only like the line of sight.

I'm sorry, but I still do not understand. I cannot picture what this angle would look like.

Why do I not simply take the reverse tangent of the x and y components to find the angle?
tan^-1 (y/x)

 

[lanedance]

please re-read the posts

you need to find two angles & the difference between them

 

[Gold3nlily]

please re-read the posts

you need to find two angles & the difference between them

Thank you for your help. "The line of sight" and "the relative velocity" are the same vector. Since they overlap, there is no angle between them and the answer is zero. I do not think this is a very fair question at all. It is tricky and it is not asked very clearly. I do appreciate you being patient with me though. Thank you again.

 

[lanedance]

no worries, it always pays to read the question carefully & have a think about what it is actually asking before jumping into it