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 0.200 m/s2. If the acceleration from the wind lasts for 2.50 s, find the magnitude, r, and direction, 0, of the bird's displacement during this time period. r = m theta = degree 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 0.200 m/s2, find the displacement vector r RIghtarrow, and the angle of the displacement, Theta1. Enter the components of the vector and angle below. (Assume the time interval is still 2.50 s.) Rightarrow = m i + m j Theta1 = degree.

Explanation / Answer

(a)

r = rx +ry

= rx + at^2/2 ( from teh kinmatic equation)

= ((1 mph(1609 / 3600) x 2.5) i + 0.200 x2.5^2/2 j  

= 1.117i + 0.625j

|r| = sqrt(1.117^2 + 0.625^2) = 1.27 m

theta = tan-1(0.625/1.117) = 29.22 degrees

2. r = ((1mph (1609 / 3600) 2.50) i + [(0.2 x2.50^2/2) x (cos38i +sin38 j )]

r   =  1.117i +0.788 i+ 0.625j+0.384 j

=1.905 i +1.009 j

theta = tan-1(1.009/1.905) = 27.9 degrees

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