Question

An unsuspecting bird is coasting along in an easterly direction at 1.00 mph when a strong wind from the south imparts a constant acceleration of 0.300 m/s^2. If the acceleration from the wind lasts for 2.30 s, find the magnitude, r, and direction, theta, of the bird's displacement during this time period. r = Number m theta = Number degree 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 44.0 degree angle to the original direction of motion. lf the magnitude of the acceleration is 0.400 m/s^2, find the displacement vector r^vector, and the angle of the displacement, theta_1. Enter the components of the vector and angle below. (Assume the time interval is still 2.30 s.) r^vector = Number m i^circ + Number m theta_I = Number degree

Answer 1

Velocity of bird is 1mph u = lmph [1609m/1 mile][1h/3600s] = 0.44694 m/s by using kinematic equation r_1 = (ut)i = (0.44694)(2.30)j = (1.0279m)j The displacement of biral along south r_2 = 1/2 at^2 (j) = 1/2(0.300)^2 (2.30)^2 (j) = 0.7935j displacement of the bird is r = r_1 + r_2 = (1.0279)i + (0.7935)j magnitude is = squareroot (1.0279)^2 + (0.7935)^2 = 1.298m direction of motion of bird theta = tan^- 1 [0.7935/1.0279] theta = 37.66 degree \r\n acceleration of the bird is a = acos 44 i + a sin 44 j = (0.400)cos 44 i + (0.400)sin 44 j = (0.2877)i + (0.2778)j a_x = 0.2877i, a_y = 0.2778j The displacement of the bird along hoprizontal direction is r_1 = [ut + 1/2a_xt^2]i r_1 = [(0.44694)(2.30) + 1/2(0.2877)(2.30)^2]j = (1.788928m)i The displacement of the bird along vertical dirwction is r_2 = [0]t + 1/2 a_y t^2j = [1/2 (0.2778(2.30)^2)]j = (0.73478m)j magnitude = squareroot (1.788928)^2 + (0.73478)^2 = 1.93m direction theta_f = tan^- 1[0.73478/1.788928]