An unsuspecting bird is coasting along in an easterly direction at 1.00 mph when

Question

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 acceleration from the wind lasts for 3.20 s, find the magnitude, r, and direction, , of the bird's displacement during this time period. (HINT: assume the bird is originally travelling in the x direction and there are 1609 m in 1 mile.) Now, assume the same bird is moving along again at 1.00 mph in an easterly direction but this time the acceleration given by the wind is at a 30.0 degree angle to the original direction of motion. If the magnitude of the acceleration is 0.200 m/s2, find the displacement vector , and the angle of the displacement, 1. Enter the components of the vector and angle below. (Assume the time interval is still 3.20 s.) I need the x and y value and the degree of the angle

Explanation / Answer

1.00mph=0.44m/s

the displacement after 3.20s is

r=0.44m/s*3.20si+1/2*0.300*3.20*3.20

r=[1.40i+1.536j]m

where i and j are the units vectors east and north

magnitude=root(1.40*1.40+1.536*1.536)m=2.07m

the magnitude r=2.07m

direction=arctan(1.536/1.40)=45degrees north of east

now,

magnitude r=[(0.44*3.20+1/2*0.300*cos30.0*3.20*3.20)i+1/2*0.300*sin30.0*3.20*3.20)j]m

r=[1.49i+0.119j]m

r=root(1.49*1.49+0.119*0.119)=0.27m

direction=arctan(0.119/1.49)=0degree

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