unsuspecting bird is coasting along in an easterly direction at 1.00 mph when a
ID: 1331667 • Letter: U
Question
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 0.200 m/s2. If the acceleration from the wind lasts for 3.00 s, find the magnitude, r, and direction, 9, of the bird's displacement during this time period. 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 38.0 degree angle to the original direction of motion. If the magnitude of the acceleration is 0.500 m/s2, find the displacement vector r, and the angle of the displacement, theta 1. Enter the components of the vector and angle below. (Assume the time interval is still 3.00 s.)Explanation / Answer
Consider +x direction as East and +y direction as North
Initial speed: vi = 1.00 mph East = 1.00 i
Convert mph to m/s
vi = (1.00 i )(1609 m / 1 mile) ( 1 hr / 3600 s) =( 0.447 ) i m/s
Acceleration:
a = 0.2 m/s2 north j m/s2
Use the equation of motion:
sf = vi*t + 1/2*a*t^2
= (0.447 i + 0j m/s) ( 3.0 s) + 1/2 ( 0.2 m/s2) (3.0 s)2
= ( 1.34 i ) m + ( 0.9 j )m
Therefore, after 3.0 s, the birds displacement from the start is 1.34 m in the East direction and 0.9 m in the North direction.
You have the two legs of the triangle now. the resultant of these two,
||sf|| = sqrt(1.34^2 + 0.9)2 m = 1.61 m
direction = tan-1 ( 0.9 / 1.34 ) = 33.86o
or 33.86o North of East or 56.1o Compass
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In this given case, the initial speed:
vi = ( 0.447 ) i m/s
and acceleration
a = 0.200 m/s2 . 380 above/below East (x-axis, i-axis) =
= 0.200 m/s2 * (cos(38 )i + sin(38)j)
= 0.157 i + 0.123 j m/s2
Again, sf = vi*t + 1/2*a*t^2
sf = ( 0.447 i ) (3.0 s) + 1/2 ( 0.157 i + 0.123 j) (3.0 s)^2
= ( 1.34i ) + (0.7065 i + 0.5535 j m)
= (2.05)i + (0.5535 ) j m
Direction:
angle = tan-1 ( 0.5535 / 2.05 ) = 15.1o
or 15.1o deg North of East or 74.8o Compass
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