# physics help?

A swan is flying at a speed of 17.5 m/s. There is a wind blowing

from the East at 12.5 m/s.

a) If the swan pointed due South, what would be the magnitude and direction of its

velocity relative to the ground.

b) If the swan wishes to travel due South, what would be the magnitude and

direction of its velocity relative to the ground.

c) If the swan travels due South as in part b, what will be its displacement after 8.5

### 4 Answers

- NCSLv 75 months agoFavourite answer
Let's assume that 17.5 m/s is the air speed of the swan (as opposed to the ground speed).

a) Since the air speed and wind speed vectors are perpendicular, use the Pythagorean theorem:

mag v = √(17.5² + 12.5²) m/s = 21.5 m/s

and the direction is

Θ = arctan(12.5/17.5) = 35.5º W of South

or 54.5º S of W

which is 234º ccw from "east"

b) If the swan wants to travel south, it must counter the wind. Measuring the angle from south,

17.5m/s * sinΘ = 12.5 m/s

Θ = arcsin(12.5/17.5) = 45.6º E of South

which is 44.4º S of E, or

314º ccw from "east"

The wind having been effectively canceled, the ground speed is

v = 17.5m/s * cos45.6º = 12.2 m/s

c) After 8.5 what?

After 8.5 s, d = 104 m

After 8.5 min, 6250 m

After 8.5 h, 375 km

Hope this helps!

- Old Science GuyLv 75 months ago
see diagrams for solutions

a

(from east means toward west)

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please consider giving a best answer.

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- oubaasLv 75 months ago
a)

Vx = -12.5 m/sec

Vy = -17.5 m/sec

module of V = √ 12,5^2+17,5^2 = 21,51 m/sec

heading = arctan (-12.5 / -17.5) = 35.54 ° WoS

b)

module of V' = 17.5 m/sec

V'x = 12.5 m/sec

V'y = √ -(12,5^2)+17,5^2 = -12.25 m/sec

heading = arctan (V'y/V'x) = arctan(-12.25/12.5) = -44.4° (44.4° SoE)

c)

d = V'y*t = 12.25*8.5 = 104.1 m