find the displacement vector and the angle of the displacement of a bird's travel

Question

assume a bird is moving along at 3.00 mph in an easterly direction, the acceleration given by the wind is at a 31.0° angle to the original direction of motion, measured counterclockwise from the easterly direction. If the magnitude of the acceleration is 0.200 m/s², find the displacement vector and the angle of the displacement 𝜃.

Assume the time interval is 3.90 s.

Answer 1

3.00 mi/hr = 1.34 m/s

the acceleration of 0.200 m/s² in the direction 31° ccw of East has two components:

The East component is a_E = (0.200)cos(31)

The North component is a_N = (0.200)sin(31)

Now solve two displacement equations, at t = 3.90 s, for the East and North components of displacement:

East component: x_E = v_0Et + ½a_Et² = (1.34)(3.9) +½(0.200)cos(31)(3.9)² = 6.53 m

North component: x_N = v_0Nt + ½a_Nt² = (0)(3.9) +½(0.200)sin(31)(3.9)² = 0.783 m

The magnitude of the displacement vector is √[(x_E)² + (x_N)²] = √[(6.53)² + (0.783)²] =

6.58 m

The direction of the displacement vector is tan^–1(x_N/x_E) = tan^–1(0.783/6.53) =

6.84° ccw of East