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# A pedestrian walks east at a rate of 6 km/hr. The wind pushes him northwest at a rate of 13 km/hr. Find the direction and magnitude?

of resultant vector.

### 3 Answers

- Jeff AaronLv 71 year agoFavorite Answer
It's unlikely that the wind would have much effect, due to friction between the pedestrian and the surface (e.g. sidewalk), and that friction is generally necessary for the pedestrian to walk in the first place. However, if we imagine that this is an airplane, not a pedestrian:

c^2 = a^2 + b^2 - 2ab*cos(C)

c^2 = 6^2 + 13^2 - 2*6*13*cos(45 degrees)

c^2 = 36 + 169 - 156*sqrt(0.5)

c = sqrt(205 - 156*sqrt(0.5)), note that a distance (e.g. c) can't be negative

c =~ 9.3982647921954007091260564482919

sin(B)/b = sin(C)/c

sin(B)/13 = sin(45 degrees)/sqrt(205 - 156*sqrt(0.5))

sin(B) = 13*sqrt(0.5)/sqrt(205 - 156*sqrt(0.5))

B = arcsin(13*sqrt(0.5)/sqrt(205 - 156*sqrt(0.5)))

Since 90 degrees <= B <= 180 degrees, as the "airplane" will end up travelling approximately northwest:

B =~ 109.151 degrees

This means that we start pointing east and rotate approximately 109.151 degrees to the left, so the final direction is about 19.151 degrees west of north.

Note: Since the wind speed is greater than the "airplane's" speed, the "airplane" shouldn't have "taken off" in the first place! It's too dangerous!

- BryceLv 71 year ago
6cos0°i + 6sin0°j + 13cos135°i + 13sin135°j

≈ -3.19i + 9.19j

magnitude= √(-3.19² + 9.19²)≈ 9.73 km/hr

Θ= arctan(9.19/-3.19)≈ -70.9°

180 - 70.9= 109.1° or N19.1°W (19.1° west of north)

- billrussell42Lv 71 year ago
The problem is: you can't combine a wind speed with a walking rate as you don't know how that wind speed changes his walking rate and direction. If he is determined, it does not change it at all.

But I'll do a simple vector addition.

Walking, Ax = 6, Ay = 0

Wind, Bx = –13 cos 45 = –13/√2 = –9.19

By = +9.19

add the x and y components

Tx = 6–9.19 = –3.19

Ty = 9.19

T = √(3.19² + 9.19²) = 9.74 km/hr

θ = arctan (9.19/3.19) = 70.9º

from a drawing, that is angle N of E