An unsuspecting bird is coasting along in an easterly direction at 1.00 mph when a strong wind from the south imparts a constant acceleration of 0.300 m/s2. If the wind\'s acceleration lasts for 3.40 s, find the magnitude r and direction θ (measured counterclockwise from the easterly direction) of the bird\'s displacement over this time interval. (HINT: assume the bird is originally travelling in the x direction and there are 1609 m in 1 mile.)
Assume that
+ = +x = East
+ = +y = North
Part 1)
vi = 1.00 mph East = 1.00 + 0 mph
Convert mph to m/s
vi = (1.00 + 0 ) mph x (1609 / 1 mile) x (1 hr / 3600 s) = 0.447 + 0 m/s
a = 0.300 m/s2 From the South = 0.300 m/s2 North = 0 + 0.300 m/s2
Equation of motion:
u = 0
r = 2.30 m
Direction North of East or 41.22o compass
Part 2
vi = (0.447 + 0 ) m/s
a = 0.500 m/s2 34.0 deg above or below East (x-axis, y-axis) = 0.500 m/s2 x (cos(34.0) + sin (34.0) ) =
a = (0.414 + 0.279 ) m/s2
cartesian or 12.04o North of East or 77.96o Compass
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