An unsuspecting bird coasting along in an easterly direction at 1.00 mph when a

Question

An unsuspecting bird coasting along in an easterly direction at 1.00 mph when a strong wind from the south impartsa constant acceleration of 0.200 m/s2. If the acceleration from the wind lasts 2.30 s, find the magnitude, r, and a 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 = ? Direction = ?

Now assume the bird is moving along again at 1.00 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 0.200 m/s^2, find the displacement vector r->, and the angle of the displacement theta1. (Assume the time interval is still 2.30s)

r = number m i + number m j

Direction =?

Explanation / Answer

1mph = 0.44704 m/s

Part 1)
u = 0.44704 i + 0 j

a = 0 i + 0.2 j
t = 2.3 sec

v = u + at
vy = 0 + 0.2*2.3 = 4.6

sy = ut + 1/2*a*t^2
sy = 0 + (1/2)*0.2*(2.3)^2
sy = 0.529

sx = 0.44704*2.3 = 1.0282

r = 1.0282 i + 4.6 j
Magnitude = ((1.0282)^2 + (4.6)^2)^(1/2) = 4.714
Angle = arctan(4.6/1.0282) = 77.4 degrees north of east

Part 2)
u = 0.44704 i + 0 j
a = 0.2cos38 i + 0.2*sin38 j = 0.1576 i + 0.1231 j
t = 2.3 sec

v = u + at
vy = 0 + 0.1231*2.3 = 0.28313

sy = ut + 1/2*a*t^2
sy = 0 + (1/2)*0.1231*(2.3)^2
sy = 0.3256

sx = ut + 1/2*a*t^2
sx = 0.44704*2.3 + 1/2*(0.1576)*(2.3)^2
sx = 1.445

R = 1.445 i + 0.28313 j
Magnitude = ((1.445)^2 + (0.28313)^2)^(1/2) =1.4724
Angle = arctan(0.28313/1.445) = 11.09 degrees north of east

Please give a thumps up if the answers are correct.

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