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A pet-store supply truck moves at 25.0 m/s north along a highway. Inside, a dog moves at 1.75 m/s at an angle of 35.0 degrees east of north. What is the velocity of the dog relative to the road?

I did all of my math, but I somehow got a negative number for the resultant vector, but if I'm not mistaken, the resultant vector is going in a positive direction, so something is not adding up... I guess the only thing I need help with is figuring out what angle I'm supposed to use for the Law of Cosine formula?

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SaniShahbaz

Answer:

Velocity of the dog relative to the road = 26.43 m/s

Explanation:

"For better understanding picture is attached with the answer"

Let us assume Velocity of the truck is V

And Velocity of the dog is u  

V = 25 m/s

u = 1.75 m/s

Angle = 35° East to north (East of north mean reference axis is north and the angle is between north and east axis with respect to north axis)

Let

Horizontal component of dog velocity is towards north = [tex]u_{x}[/tex] = ucos35°

Vertical component of dog velocity is towards east = [tex]u_{y}[/tex] = usin35°

[tex]u_{x}[/tex] = ucos35°

[tex]u_{x}[/tex] = 1.75cos35°

[tex]u_{x}[/tex] = 1.43 m/s

[tex]u_{y}[/tex] = usin35°

[tex]u_{y}[/tex] = 1.75sin35°

[tex]u_{y}[/tex] = 1.00 m/s

Velocity of  the truck is towards north and we assume horizontal component of dog  velocity is towards north which is equal to 1.43 m/s.  So, to calculate the  velocity of the dog relative to the road, we add up both of these velocities  and this velocity is along the road.  One component of dog velocity is  towards east,  perpendicular to the  road which is equal to 1.00 m/s.

Velocity of the dog relative to the road = 25 + 1.43

Velocity of the dog relative to the road = 26.43 m/s

Angle = tan⁻¹(1.00/26.43)

Angle = 2.17° East to north

Ver imagen SaniShahbaz

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