Vb = 4mi/h * 1609m/mi * 1h/3600s = 1.79
m/s.
Vw = a*t = 0.30 * 2.7 = 0.81 m/s.
Vr. = sqrt(Vb^2+Vw^2) = 1.96 m/s. = Resultant velocity.
D = Vr*t = 1.96*2.7 = 5.29 m.
Tan A = Y/X = Vw/Vb = 0.81/1.79=0.45251
A = 24.3o = Direction.
An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the south imparts a constant acceleration of 0.300 m/s2. If the acceleration from the wind lasts for 2.70 s, find the magnitude, r, and direction, θ, 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.)
2 answers
Vb = 4mi/h * 1609m/mi * 1h/3600s = 1.79
m/s
To solve the problem, you need to create a formula with a position as a function of time for the bird(x) and the wind(y), and then use R = sqrt(Rx^2 + Ry^2) to find the resultant magnitude and the inverse tangent of y/x
R = Ri(initial) + Vi*t + 1/2(Ri)t^2
Rx = 0 + 1.79(2.7) + 1/2(0)(2.7)^2 = 4.833
Vi for the wind is 0 because it is with respect to the bird for the y coordinate
Ry = 0 + 0(3.7) + 1/2(0.3)(3.7)^2 = 2.054
R= sqrt(4.833^2 + 2.054^2) = 5.251 meters
A = arctan(2.054/4.833) = 23.025 degrees
m/s
To solve the problem, you need to create a formula with a position as a function of time for the bird(x) and the wind(y), and then use R = sqrt(Rx^2 + Ry^2) to find the resultant magnitude and the inverse tangent of y/x
R = Ri(initial) + Vi*t + 1/2(Ri)t^2
Rx = 0 + 1.79(2.7) + 1/2(0)(2.7)^2 = 4.833
Vi for the wind is 0 because it is with respect to the bird for the y coordinate
Ry = 0 + 0(3.7) + 1/2(0.3)(3.7)^2 = 2.054
R= sqrt(4.833^2 + 2.054^2) = 5.251 meters
A = arctan(2.054/4.833) = 23.025 degrees
0 answers