An unsuspecting bird is coasting along in an easterly direction at 2.00 mph when

Question

An unsuspecting bird is coasting along in an easterly direction at 2.00 mph when a strong wind from the south imparts a constant acceleration of 0.200 m/s2. If the acceleration from the wind lasts for 3.90 s, find the magnitude, r, and 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.)

a) r = ? Need in meters

Now, assume the same bird is moving along again at 2.00 mph in an easterly direction but this time the acceleration given by the wind is at a 39.0 degree angle to the original direction of motion. If the magnitude of the acceleration is 0.400 m/s2, find the displacement vector , and the angle of the displacement, ?1. Enter the components of the vector and angle below. (Assume the time interval is still 3.90 s.)

Explanation / Answer

Speed = 2.00 mph = 0.894meter/s

Vx = 0.894 m/s
Vy = 0

ax = 0
ay = 0.2 m/s^2


Displacement in x direction = vx * t
rx= 0.894 * 3.9 m
rx = 3.49 m

Displacement in y direction = vy * t + 0.5 * a *t^2
ry = 0 + 0.5 * 0.2 * 3.9^2
ry = 1.52
r = sqrt(rx^2 + ry^2)
r =sqrt(3.49^2 + 1.52^2)
r = 3.81 m

= tan^-1(ry/rx)
= tan^-1(1.52/3.49)
= 23.53 0

b)

Speed = 2.00 mph = 0.894meter/s

Vx = 0.894 m/s
Vy = 0

ax = 0.4 * cos(39) m/s^2
ax = 0.311 m/s^2

ay = 0.4 * sin(39) m/s^2
ay = 0.252 m/s^2

Displacement in x direction = vx * t + 0.5 *ax*t^2
rx= 0.894 * 3.9 m + 0.5 * 0.311 * 3.9^2
rx = 5.85 m

Displacement in y direction = vy * t + 0.5 * a *t^2
ry = 0 + 0.5 * 0.252 * 3.9^2
ry = 1.92 m

r = 5.85 i m + 1.92 j m

= tan^-1(ry/rx)
= tan^-1(1.92/5.85)
= 18.17o

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