Find the angle of the second displacement (measured counterclockwise from the positive x axis). Answer in units of ◦ .

Question

A particle undergoes two displacements. The first has a magnitude of 11 m and makes an angle of 63 ◦ with the positive x axis. The result after the two displacements is 6.8 m directed at an angle of 109◦ to the positive x axis using counterclockwise as the positive angular direction. Find the angle of the second displacement (measured counterclockwise from the positive x axis). Answer in units of ◦ .Find the angle of the second displacement (measured counterclockwise from the positive x axis). Answer in units of ◦ .

Answer 1

This problem is solved by addition of vectors representing two displacement AB and CD forming a result PQ.

In polar notation Vector AB = 11 angle 63 degrees

Vector PQ = 6.8 angle 109 degrees

Vector CD = M angle ψ

We can write i, and j components of AB = 11 cos 63 i + 11 sin 63 j = 11 x 0.454 i + 11 x 0.891 j

=4.994i + 9.8j

Component i and j of result vector PQ = 6.8 cos 109 i + 6.8 sin 109j = 6.8 x (-0.326) i + 6.8 x 0.941 j

= -2.217i +6.433j

From addition rules of vectors vector AB + Vector CD = Vector PQ

Writing components 4.994 i + 9.8 j + M angle ψ = -2.217 i + 6.433 j

Simplify and solve for M angle ψ = (-2.217 - 4.994) i + (6.433-9.8)j = -7.221 i - 3.367 j

Transform i and j components of CD to polar form magnitude = 7.958 Angle ψ = arc tan (-3.367/-7.221) = 25 degrees in third quadrant = 180 + 25 = 205 degrees

I hope this helps.

Shailesh (Sky) Kadakia, Expert Math and Science Tutor with WYZANT