Two highways intersect as shown in the figure. At the instant shown, a police ca

Question

Two highways intersect as shown in the figure. At the instant shown, a police car P is distance dP = 890 m from the intersection and moving at speed vP = 89 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 the figure, what is the angle between the velocity found in (a) and (b) and the line of sight between the two cars?

Please explain your answer

Explanation / Answer

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=75km/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? I have both of the components and I am looking for the angle. I plug them into the equation: (/) = tan^-1 (-51/75) and I get -38.1 degrees.

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