physics question below?

Question

A duck has a mass of 2.5 kg. As the duck paddles, a force of 0.10 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.18 N in a direction of 46° south of east. When these forces begin to act, the velocity of the duck is 0.08 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 2.5 s while the forces are acting.

Answer 1

Start with some equations:
F = ma
d = vt + 1/2at^2

Step 1: Convert the forces given into to 2 forces: one acting east, and one acting south.
F1 = .10N E
F2 = .18N 46° S of E

To convert F2, construct a triangle, such that the angle is 46°, the adjacent side is the eastern force, and the opposite side is the southern force.
F2E = .18N * cos 46° = 0.12504 N
F2S = .18N * sin 46° = 0.12948 N

Now, add F1 and F2E to get the eastern Force:
FE = F1 + F2E = .10 + .12504 = .22504 N
FS = 0.12948 N

Step 2: Find the Southern displacement:
F = ma
a = F/m = 0.12948 / 2.5 = 0.051792 m/s
d = vt + 1/2at^2
vt = 0
dS = 1/2 at^2 = .5 * 0.051792 * 2.5^2 = 0.16185 m

Step 3: Find the Eastern displacement:
F = ma
a = F/m = .22504 / 2.5 = 0.090016 m/s
d = vt + 1/2at^2
dE = .08 * 2.5 + 1/2 * 0.090016 * 2.5^2 = .2 + 0.2813 = 0.4813 m

Step 3: Given the Eastern and Southern displacement, use the Pythagorean Theorem to find the total displacement, and use the tangent to find the angle.
d^2 = dS^2 + dE^2
dS = 0.16185 m
dE = 0.4813 m
d^2 = 0.23164969 + 0.0261954225 = 0.2578451125
d = 0.50778 m

tan a = opp / adj = dS / dE
tan a = 0.16185 / 0.4813
tan a = 0.33628
a = 18.58° S of E.

Total displacement is 0.50778 m at 18.58° S of E. (solution!)