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/s2. If the acceleration from the wind lasts for 2.00 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.)

Answer: r=_________m

Answer: =_________degrees

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 32.0 degree angle to the original direction of motion. If the magnitude of the acceleration is 0.200 m/s2, find the displacement vector , and the angle of the displacement, 1. Enter the components of the vector and angle below. (Assume the time interval is still 2.00 s.)

Answer: r=_________m i+ __________m j

Answer: =_________degrees

An unsuspecting bird is coasting along in an easterly direction at 3.00 mph when a strong wind from the Map south imparts a constant acceleration of 0.300 m/s f the acceleration from the wind lasts for 2.00 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 m Number 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 32.0 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, o1. Enter the components of the vector and angle below. (Assume the time interval is still 2.00 s.) Number Number Number

Explanation / Answer

I prefer doing vectors in the i+j format.

+i = +x = East

+j = +y = North

Part 1)

vi = 3.00 mph East = 3.00i + 0j mph

Convert mph to m/s

vi = (3.00i + 0j) mph * (1609 m / 1 mile) * ( 1 hr / 3600 s) = 1.34i + 0j m/s

a = 0.300 m/s^2 FROM THE SOUTH = 0.300 m/s^2 North = 0i + 0.300j m/s^2

Equation of motion:

sf = si + vi*t + 1/2*a*t^2

Assume si = 0

sf = vi*t + 1/2*a*t^2

sf = (1.34i + 0j m/s)*2 s + 1/2 * (0i + 0.300j m/s^2) * (2 s)^2

sf = (2.68i + 0j m) + (0+0.6j m)

sf = (2.68+0)i + (0i+0.6)j m

sf = 2.68i + 0.6j m

So, after 2 s, the birds displacement from the start is 2.68m in the East direction and 0.6m in the North direction. You have the two legs of the triangle now.

||sf|| = sqrt(2.68^2 + 0.6^2) m = 2.75 m

direction = atan(j/i) = 12.6 deg Cartesian or 12.6 deg North of East

Part 2)

vi = 1.34i + 0j m/s

a = 0.200 m/s^2 32.0 deg above/below East (x-axis, i-axis) = 0.200 m/s^2 * (cos(32.0)i + sin(32.0)j) = 0.17i + 0.106j m/s^2

Again, sf = vi*t + 1/2*a*t^2

sf = (1.34i + 0j m/s)*2 s + 1/2 * (0.17i + 0.106j m/s^2) * (2 s)^2

sf = (2.68i + 0j m) + (0.34i+0.212j m)

sf =3.02i + 0.212j m

angle = atan(j/i) = atan(0.212/3.02) = 4.02 deg Cartesian or 4.02 deg North of

Actually, since they didn't specfiy which direction the wind was blow, North of East or South of East, then just assume the direction is 4.02 deg off the x-axis, so it could be +4.02 deg or -4.02 deg. Take your pick.

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