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
- 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
(from east means toward west)
When you get a good response,
please consider giving a best answer.
This is the only reward we get.
- oubaasLv 75 months ago
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
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)
d = V'y*t = 12.25*8.5 = 104.1 m
- RealProLv 75 months ago