Posted by **karl** on Thursday, August 7, 2014 at 6:09am.

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.

- trigonometry -
**Reiny**, Thursday, August 7, 2014 at 7:46am
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

- trigonometry -
**Henry**, Thursday, August 7, 2014 at 8:10pm
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.

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