Automobile A is traveling east at 36 kmh As A passes the int

Automobile A is traveling east at 36 km/h. As A passes the intersection shown, automobile B starts from rest 35 m north of the intersection and moves south with a constant acceleration of 1.2 m/s^2. Determine the position, velocity, and acceleration of B relative to A at 5 s after A crosses the intersection.

Solution

Relative to the observer who is at rest at the intersection.
A has moved a distance of 10*5 = 50 m
B is at a distance of 35 – 0.5*1.2*5*5 = 20 m from the observer.
A’s velocity after 5 second is the same as 10 m/s
B’s velocity after 5second is at = 1.2*5 = 6 m/s toward south .
A’s acceleration after 5 second is zero
B’s acceleration after 5second is = 1.2 mn/s^2
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With respect to A, B is at a distance of (50^2 + 20^2) = 53.85 m from him.
tan = 20 /30 => = 33.69°

That is the position with respect to A, B is at angle of (90 - 33.69) = 56.31°from the positive x direction and in counter clockwise.
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With respect to A, B’s velocity is (10^2 + 6^2) = 11.66 m/s

tan = - 6 /10 => = -30.96° (minus indicates below the x axis
With respect to A, B’s acceleration is 1.2 mn/s^2 toward south