Question
HELP second parts not right
An unsuspecting bird is coasting along in an easterly direction at 1.00 mph when a strong south imparts a constant acceleration of 0.400 m/s2. If the acceleration from the wind lasts Tor 2.20 s, find the magnitude, r, and direction, theta, 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.) Now, assume the same 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 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, theta1. Enter the components of the vector and angle below. (Assume the time interval is still 2.20 s.)
Explanation / Answer
initially u = 1mph = 1 x1609 / 3600 = 0.447 m/s
so velocity vector v = 0.447 i m/s
acceleration given by wind = 0.5 m/s2 ....at an angle of 46 degree
so acce. vector a = 0.5cos46i + 0.5sin46j m/s2
so displacement vector :
r = (v_xt + a_xt^2/2)i + (v_y t + a_y t^2/2)j
= (0.447 x 2.20 + 0.5cos46 x 2.20^2/2)i + (0 + 0.5sin46 x 2.20^2/2)j
= 1.82i + 0.870j m ............................ANs
angle = tan-1(-0.870 / 1.82) = 25.55 degrees