Two friends, Barbara and Neil, are out rollerblading. With respect to the ground, Barbara is skating due south at a speed of 4.6 m/s. Neil is in front of her. With respect to the ground, Neil is skating due west at a speed of 4.4 m/s. Find Neil's velocity (a) magnitude and (b) direction relative to due west, as seen by Barbara.

a) the magnitude is 6.37 m/s

b) the direction relative to due west, as seen by Barbara is 46.37° north of west

Explanation:

Given the data in the question;

Barbara is skating due south at a speed of 4.6 m/s

Neil is skating due west at a speed of 4.4 m/s

so

→ = 4.6 (-j^)

→ = 4.4 (-i^)

→ = → - →  = 4.4 (-i^) - 4.6 (-j^)

= -4.4i^ + 4.6j^

a) magnitude

magnitude |(→ )| = √( (-4.4)² + (4.6)²)

= √( 19.36 + 21.16)

= √40.52

= 6.37 m/s

Therefore, the magnitude is 6.37 m/s

b) direction relative to due west, as seen by Barbara.

using SOH CAH TOA

tan = opp / adj

so direction will be;

∅ = tan⁻¹( 4.6 / -4.4)

∅ = tan⁻¹( - 1.0454545 )

∅ = - 46.37°

Therefore, the direction relative to due west, as seen by Barbara is 46.37° north of west

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

where’s the list ? the question is incomplete