John and Mary are skating at an ice rink. John skates at a constant speed of 6.7 m/s, with respect to the ice surface, directly south. Mary skates at a constant speed of 11.5 m/s in a direction of 34° west of south, with respect to the ice. What are the magnitude and direction of Mary's velocity with respect to John due south?
V = 11.5m/s[236o] - 6.7m/s[270o]
V = 11.5*Cos236 + 11.5*sin236 - (-6.7i) = -6.43 - 9.53i + 6.7i = -6.43 - 2.83i
Tan Ar = Y/X = -2.83/-6.43 = 0.44074
Ar = 23.8o = Reference angle.
A = 23.8 + 180 = 203.8o, CCW = 66.2o W.
of S. = Direction.
V = X/Cos2o3.8 = -6.43/-.915 = 7.03 m/s.
= Magnitude.
To find the magnitude and direction of Mary's velocity with respect to John due south, we can use vector addition.
First, let's break down Mary's velocity into its north and east components:
Mary's velocity (in m/s) = 11.5 m/s at an angle of 34° west of south.
North component of Mary's velocity = 11.5 m/s * sin(34°) ≈ 6.31 m/s (positive north)
East component of Mary's velocity = 11.5 m/s * cos(34°) ≈ 9.57 m/s (positive east)
Since John is skating directly south, his velocity does not have any east or north components. Therefore, his velocity vector can be represented as 6.7 m/s * (-1) south.
Now, to find Mary's velocity with respect to John due south, we need to subtract John's velocity from Mary's velocity. Since they both have south as the common direction, we only need to consider the magnitudes.
Magnitude of Mary's velocity with respect to John due south = |Magnitude of Mary's velocity - Magnitude of John's velocity|
= |6.7 m/s - (-1 m/s)|
= |6.7 + 1|
= |7.7|
= 7.7 m/s
Therefore, the magnitude of Mary's velocity with respect to John due south is 7.7 m/s.
The direction of Mary's velocity with respect to John due south is still south since both velocities are aligned in the same direction.
So, the magnitude of Mary's velocity with respect to John due south is 7.7 m/s, and the direction is due south.