An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when

Question

An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the Mapd: south imparts a constant acceleration of 0.300 m/s f the acceleration from the wind lasts for 3.20 s, find the magnitude, r, and direction, 6, 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.) Number Number Now, assume the same bird is moving along again at 4.00 mph in an easterly direction but this time the acceleration given by the wind is at a 31.0 degree angle to the original direction of motion. If the magnitude of the acceleration is 0.200 m/s2, find the displacement vector r, and the angle of the displacement, 01. Enter the components of the vector and angle below. (Assume the time interval is still 3.20 s.) Number Number ml Number

Explanation / Answer

Birds speed is = 4*1609/3600 = 1.7877 m/s in 3.2 s x = v*t = 6.1102 m y = 0.5*0.3*3.2^2 = 1.536 m r=Sqrt(x^2 +y^2) = 6.3 m tan theta =1.536/6.112 theta = 14.1 degrees Now ax =0.2*cos31 = 0.1714 m/s^2 x =v*t + 0.5*ax*t^2 = 6.5983 mi ay = 0.2*sin31 =0.103 m/s^2 y = 0.5*a*t^2 = 0.527 m j so r = 6.6 i + 0.527 j theta = 4.57 degrees Please rate my last submissions also...

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