An unsuspecting bird is coasting along in an easterly direction at 2.00 mph when

Question

An unsuspecting bird is coasting along in an easterly direction at 2.00 mph when a strong wind from the south imparts a constant acceleration of 0.300 m/s^2. If the wind's acceleration lasts for 2.30 s, find the magnitude r and direction theta (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. r = theta = Now, assume the same bird is moving along again at 2.00 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/s^2, find the displacement vector r^vector, and the angle of the displacement, theta_1. Enter the components of the vector and angle below. (Assume the time interval is still 2.30 s.) r^vector = i^cap + j^cap theta_1 =

Explanation / Answer

v0 = 2 mph i

v0 = (2 x 1609 / 3600 m/s) i = 0.89 i

a = 0.300 m/s^2 (-j)

r = v0 t + a t^2 /2

r = 2.06i - 0.79 j m

magnitude = sqrt(2.06^2 + 0.79^2) = 2.20 ....Ans

direction = 360 - tan^-1(0.79/2.06) =339 deg ......Ans

a = 0.300 (cos38i - sin38j) = 0.236i - 0.185j m/s^2

r = (0.89i)(2.30) + (0.236i - 0.185j)(2.30^2 / 2)

r = 2.05i + 0.624i - 0.489j

r = 2.67i - 0.49 j

r = (2.67 m) i + ( - 0.49 m) j

direction = 360 - tan^-1(0.49 / 2.67 ) = 349.6 deg

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