Question

A student bikes to school by traveling first d_N = 1.10 miles north, then d_W = 0.500 miles west, and finally d_S = 0.200 miles south.

Part A

 If a bird were to start out from the origin (where the student starts) and fly directly (in a straight line) to the school, what distance d_b would the bird cover?

Part B 

 Let the vector d_N be the displacement vector corresponding to the 1st leg of the trip. Express d_N in component form.

Part C 

 For the 2nd leg (d_W):

Part D 

 For the 3rd leg (d_S):

Part E 

 The displacement vector for the bird, d_b can be written as d_N + d_W + d_S. Express this vector in component form:

Part F 

 Find φ, the angle North of West of the path followed by the bird.

Answer 1

Part A Answer

d_b = SQRT(d_x² + d_y²) = SQRT((-0.500)² + (1.10 – 0.200)²)

d_b = SQRT(0.25 + 0.81)

d_b = SQRT(1.06) or 1.0296

Part B Answer

d_N = (0.0, 1.1)

Part C Answer

d_W = (-0.5, 0.0)

Part D  Answer

d_S = (0.0, -0.2)

Part E Answer

Just add the north and south directions together (1.1 – 0.2 = 0.9)

d_b = (-0.5, 0.9)

Part F Answer

Remember SOHCAHTOA. In this case, Tan(φ) = Opposite / Adjacent. North of west means y is the opposite, x the adjacent. So:

Tan(φ) = 0.9 / 0.5
Tan(φ) = 1.8

φ = Tan^-1(1.8)

φ = 60.9°