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 Map south imparts a constant acceleration of 0.300 m/s. If the wind's acceleration lasts for 3.40 s, find the magnitude r and direction measured counterdock se from the easterly drect on of the birds displacement over this time interval. (HINT: assume the bird is originally travelling in the +x direction and there are 1609 m in 1 mile.) Number Im Number = Now, assume the same bird is moving along acceleration given by the wind is at a 37.0 degree angle to the original direction of motion. If the magnitude again at 2.00 mph in an easterly direction but this time the of the acceleration is 0.500 m/s2, find the displacement vector r, and the angle of the displacement, 1. Enter the components of the vector and angle below. (Assume the time interval is still 3.40 s.) Number Number Number o-revious "Check Answer 0 Next Exit Hint

Explanation / Answer

v0 = 2 mph = 2 x 1609 m / 3600s = 0.894 m/s i

a = 0.30 j m/s^2

t = 3.40 sec

r = v0 t + a t^2 / 2

r = (0.894 x 3.40)i + (0.30 x 3.40^2 / 2)j

r = 3.04 i + 1.73 j

|r | = sqrt(3.04^2 + 1.73^2) = 3.50 m .......Ans

theta = tan^-1(1.73 / 3.04) = 30 deg .....Ans

a = 0.5 ( cos37i + sin37 j) = 0.40 i + 0.3 j m/s^2

r = v0 t + a t^2 /2 = 5.35 i + 1.73 j   

r = ( 5.35 m ) i + (1.73 m ) j

theta = tan^-1(1.73/5.35) = 18 deg

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