A barge is being towed south at the rate 2 km/hr. A man on the deck walks from west to east at the rate 4ft/s. Find the magnitude and direction of the man's actual velocity.

Answer 1

1 km = appr 3280.84 feet
2 km/h = 2(3280.84/3600) ft/s
= 1.82269 ft/s

let r be the resultant velocity
r^2 = 2^2 + 1.82269^2
r = √7.32219..
= appr 2.7 ft/s

tanØ = 1.82269/2 , where Ø is in my right-angled triangle
Ø = appr 42.34°

his direction is S 42.34° E

or by vectors
vector r = 2(cos270 , sin270) + 1.82269(cos0 , sin0)
= 2(0, -1) + 1.82269(1,0)
= (0,-2) + (1.82269,0)
= (1.82269,-2)

magnitude = √(1.82269^2 + (-2)^2)
= 2.7 as above
direction:
tan k = -2/1.82269
k = 312.34° which is the same as S 42.34 W

Answer 2

X = 4 Ft/s

Y = -2km/h = -2000m/3600s * 3.3Ft/m = -1.833 Ft/s

Tan A = Y/X = -1.833/4 = -0.45825
A = -24.62o = 24.62o South of East. =
65.38o East of South = Direction.

Magnitude = X/Cos A = 4/Cos(-24.62) =
4.4 Ft/s.