An unsuspecting bird is coasting along in an easterly direction at 3.00 mph when

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/s^2. If the acceleration from the wind lasts for 3.40 s, find the magnitude, r, and direction, theta, of the bird's displacement during this time period. 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 0.200 m/s^2, find the displacement vector r, and the angle of the displacement, theta_1. Enter the components of the vector and angle below. (Assume the time interval is still 3.40 s.)

Explanation / Answer

Horizontal velocity of Bird = 3 mph = 3* 0.447 = 1.341 m/s i^

Acceleration = 0.2 m/s^2 j^

Initial Vertical Velocity = 0

time t = 3.4 s

Displacement in Vertical Direction = u*t + 0.5at^2

S = 0 + 0.5*0.2*3.4^2

S = 1.156 m (ANSWER)

Displacement in Horizontal Direction S = ut

S = 1.346*3.4

S = 4.56 m

Net Displacement, r = sqrt(4.56^2 + 1.156^2)

Magnitude |r| = 4.704 m

Direction = tan^-1(1.156/4.56)

= 14.22 deg
-------------------------------------------------------------
theta = 47 deg

Displacement in Vertical Direction = u*t + 0.5at^2

S = 0 + 0.5*0.2*3.4^2

S = 1.156sin 47

S = 0.845 m (ANSWER)

Displacement in Horizontal Direction S = ucos theta

S = 1.151*3.4 * cos 47

S = 3.11 m

so r = 0.845 i + 3.11 j (ANSWR)

Net Displacement, r = sqrt(0.845^2 + 3.11^2)

Direction = tan^-1(0.845/3.11)

= 15.22 deg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.