Answer to Question #246606 in Mechanics | Relativity for Timothy

Question

an ice skater skates 15m southwest then 30meast and finally 45m in a direction 52.5 north of east.Find the magnitude of the displacement required to bring the skater back to her original position

Answer 1

To find the magnitude of the displacement, you need to add up the x and y components of her travel:

For 15m southwest:

sin(angle) $= \frac{D_y }{ D}$

sin(45 degrees) $= \frac{D_y }{ 15}$

Dy = 10.61 m west

Since this is a 45 degree angle, $D_x = D_y = 10.61\; m$ south as well.

30 m east is just 30 m east.

45 m at an angle of 52.5 degrees north of east:

sin(angle) $= \frac{D_y }{ D}$

sin(52.5 degrees) $= \frac{D_y }{ 45}$

Dy = 35.70 m north

cos(angle) $= \frac{D_x }{ D}$

cos(52.5 degrees) $= \frac{D_x }{ 45}$

Dx = 27.39 m east

Vertical components: (35.70 m north + 10.61 m south) = 25.09 m north.

Horizontal components: (27.39 m east + 30 m east + 10.61 m west) = 46.78 m east.

$Magnitude^2 = D_x^2 + D_y^2 \\ Magnitude^2 = 46.78^2 + 25.09^2$

Magnitude = 53.08 meters