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## Homework Statement

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.

## Homework Equations

F = ma

v

^{2}-v

_{0}

^{2}= 2a(x-x

_{0})

x = x

_{0}+v

_{0}t+1/2 at

^{2}

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/s

_{2}.

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 v

^{2}-v

_{0}

^{2}= 2a(x-x

_{0}).

I substituted: .38

^{2}-.17

^{2}= 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.