An unsuspecting bird is coasting along in an easterly direction at 3.00 mph when

Question

An unsuspecting bird is coasting along in an easterly direction at 3.00 mph when a strong wind from the south imparts a constant acceleration of 0.300 m/s^2. If the acceleration from the wind lasts for 2.90 s, find the magnitude, r, and direction, theta, of the bird's displacement during this time period. R=Number 3.98 m theta=Number 12.61 degree Now, assume the same bird is moving along again at 3.00 mph in an easterly direction but this time the acceleration given by the wind is at a 44.0 degree angle to the original direction of motion. If the magnitude of the acceleration is 0.300 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 2.90 s.) r^vector=Number 5.62 m i^cap+Number 60 m j^cap theta_i=Number 6.09 degree

Explanation / Answer

a) Acceleration, a = ay = 0.300 m/s2

Displacement in x-direction, x = vxt = (3 * 1609 / 3600) * 2.90 = 3.89 m

Displacement in y-direction, y = ayt2/2 = 0.300 * 2.902 / 2 = 1.26 m

So, magnitude of displacement, r = (x2 + y2)1/2 = (3.892 + 1.262)1/2 = 4.09 m

Direction of displaement, = tan-1(y/x) = tan-1(1.26 / 3.89) =  18.0o

b) Acceleration, a = axi + ayj = (0.300 * cos44o)i + (0.300 * sin44o)j

Displacement in x-direction, x = vxt + axt2/2 = [(3 * 1609 / 3600) * 2.90] + [(0.300 * cos44o) * 2.902 / 2] = 4.80 m

Displacement in y-direction, y = ayt2/2 = (0.300 * sin44o) * 2.902 / 2 = 0.88 m

So, displacement vector, r = 4.80i + 0.88j

Direction of displaement, = tan-1(y/x) = tan-1(0.88 / 4.80) =  10.4o

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