A duck has a mass of 2.7 kg. As the duck paddles, a force of 0.13 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.20 N in a direction of 54° south of east. When these forces begin to act, the velocity of the duck is 0.13 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 2.3 s while the forces are acting.
F = ma
v2-v02 = 2a(x-x0)
x = x0+v0t+1/2 at2
v = vat
The Attempt at a Solution
First, I drew a diagram. The .13 N is a vector going from right to left. Then, the .20 N vector connects and heads south east. Because it is 54 degrees south of east, that means the interior angle where the two vectors join is 126 degrees.
I then found the results.
I found the resultant vector to be .293 N.
I then used the F=ma equation to find acceleration which I found to be .1096 m/s2.
I then plugged this in to v = v a t (v = .13 + .1096 (2.3)) to find that final velocity was .38 m/s.
I then plugged all of this into the equation v2-v02 = 2a(x-x0).
I substituted: .382-.172 = 2(.1096)x
I solved for x to be .589
I then used my diagram to find tan-1(.2/.13) to get 59 degrees as an answer.
Unfortunately, all of this was wrong.