a)An unsuspecting bird is coasting along in an easterly direction at 2.00 mph wh

Question

a)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 0.400 m/s2. If the acceleration from the wind lasts for 3.90 s, find the magnitude, r, and direction, ?, 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.

r=

theta=

b)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 50.0 degree angle to the original direction of motion. If the magnitude of the acceleration is 0.300 m/s2, find the displacement vector , and the angle of the displacement.

So pretty much I did the first half and got

r=4.6267

theta=41.15 degrees

I have no idea how to do the second part please any help is appreciated detailed answers are appreciated so I can see what you are doing thank you.

Explanation / Answer

a) u = 2mph = 2 x 1609/3600 = 0.894 m/s

in east s1 = ut = 3.49 m

south s2 = at^2/2 = 3.042 m

|r| = sqrt(s1^2 + s2^2) = 4.63 m

angle = tan-1(3.042 / 3.49) = 41.08 degrees cloackwise the east or south to east.

b) u = 2mph = 2 x 1609/3600 = 0.894i m/s

a = 0.3cos50i + 0.3sin50 j   = 0.193i + 0.230 j

s = (0.894 x 3.90 + 0.193 x 3.90^2/2)i + (0.230 x 3.90^2/2)j

s = 4.94i + 1.75 j

|s| = 5.24 m

angle = tan-1(1.75/4.94) = 19.51 degrees counter-clockwise or north to east.

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