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 0.200 m/s2. If the wind\'s acceleration lasts for 3.00 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.)

r=?

theta=?

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 47.0 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 3.00 s.)

-> ^ ^

r =number m i + number j theta=number degrees

Answer 1

given that

u = 3 mph = 1.34 m/s

t = 3s

a = 0.2 m/s²

(a)

distance in east r1 = u*t = 1.34*3 = 4.02 m

distance travelled in south r2 = 0.5*a*t² = 0.5*0.2*(3)² = 0.9 m

displacement of the bird vector r = r1 +r2 = 4.04 i + 0.9 j  

magnitude r = sqrt(r1² + r2² ) = 4.11 m

The direction of the motion of bird is

= tan^-1(0.9/4.02)

= 12.57^o

(b)

r1 = u_x*t +1/2*a_x*t²

r1 = 1.34*3 +0.5*0.2*cos47 * 9

r1 = 4.63 m

r2 = 1/2*a_y*t²

r2 = 0.5*0.2*sin47*9 = 0.658

vector r = r1 +r2 = 4.63 i + 0.658 j

magnitute r = sqrt (r1² + r2² ) = 4.67 m

Direction of the motion of bird is

= tan^-1(0.658/4.63)

= 8.08^o