- #1

Gold3nlily

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**help finding an angle please?**

## Homework Statement

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.)

## Homework Equations

Vx=Vcos(/)

Vy=Vsin(/)

V-> = (Vx)i + (Vy)j

V = (Vx^2 = vy^2)^1/2

(/) = tan^-1 (Vy/Vx)

## The Attempt at a Solution

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 am not getting any replies. Have I asked my question wrong? Is there something I should have done differently? Did I not show enough work?

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