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 0.400 m/s2. If the wind\'s acceleration lasts for 2.90 s, find the magnitude r and direction θ (measured counterclockwise from the easterly direction) of the bird\'s displacement over this time interval. (HINT: assume the bird is originally travelling in the x direction and there are 1609 m in 1 mile.)

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 46.0 degree angle to the original direction of motion. If the magnitude of the acceleration is 0.500 m/s2, find the displacement vector , and the angle of the displacement, θ1. Enter the components of the vector and angle below. (Assume the time interval is still 2.90 s.)

Answer 1

3.00 mph = 1.34 m/s

the displacement after 2.90s is
r = 1.34m/s*2.90s i + ½*0.400m/s²*(2.90s)² j = [3.89 i + 1.68 j] m
where "i" and "j" are the unit vectors "east" and "north" respectively

magnitude |r| = √(3.89² + 1.68²) m = 17.96 m ◄
direction Θ = arctan(1.68/ 3.89) = 23.5º North of East ◄

Now,
r = [(1.34*2.90 + ½*0.500*cos46.0*2.90²) i + ½*0.500*sin46*2.90² j] m
r = [5.35 i + 1.51 j] m ◄
Θ1 = arctan(1.51/5.35) = 16º ◄