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 = ?

Direction =?

Answer 1

1.00 mph = 0.447 m/s

the displacement after 2.30s is

r = 0.447m/s*2.30s i + ½*0.200m/s²*(2.30s)² j = [1.03 i + 0.529 j] m

where "i" and "j" are the unit vectors "east" and "north," respectively

magnitude |r| = √(1.03² + 0.529²) m = 1.16 m ◄

direction Θ = arctan(0.529 / 1.03) = 27.18º North of East ◄

Now,

r = [(0.447*2.30 + ½*0.200*cos38.0*2.30²) i + ½*0.200*sin38*2.30² j] m

r = [1.44 i + 0.326 j] m ◄

Θ1 = arctan(0.326/1.44) = 12.74º ◄