Finding the Displacement of a Duck Undergoing Multiple Forces

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A discussion focused on calculating the displacement of a duck under multiple forces revealed common pitfalls in applying kinematic equations. The duck, with a mass of 2.5 kg, experiences forces of 0.10 N eastward and 0.20 N at 52 degrees south of east. Participants emphasized the need to correctly compute both x and y components of acceleration and displacement. Missteps included confusing the angle of the force with the angle of displacement and neglecting to calculate the y-component of displacement. Ultimately, clarification led to a successful resolution of the problem.
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Here is the question: A duck has a mass of 2.5kg. As the duck paddles, a force of 0.10N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.20N in a direction of 52degrees south of east. When the forces begin to act, the velocity of the duck is 0.11m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 3.0s while the forces are acting.
SO, I attemted to find the acceleration for the x and y and then use the tangent-1 to find the angle and the x=Vot + 1/2a(t)squared. Again not getting the answer in the back of the book 0.78m, 21degrees south of east
Obviously I am approaching this problem wrong?! Any advice??
 
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Find the x and y components of the acceleration. Then use kinematic formulas to find the x and y components of the displacement. That sounds like what you are saying you did. Show your work so we can check it out.
 
Rx=.20N(cos 52) + .10(cos 0)=.2231N Ry= .20(sin 52) + .10(sin 0)=.1576N

Ax=.2231N/2.5kg= .08924 Ay=.1576N/2.5kg= .0252

tan-1(.0252/.08924)=15.8degrees(wrong answer)

x=.11m/s(3.0s) + 1/2(.08924m/s^2)(3.0s)^2= wrong answer!

What step am I missing or am I completely off track?
 
You are not off track, you are just making mistakes.
pinky2468 said:
Rx=.20N(cos 52) + .10(cos 0)=.2231N Ry= .20(sin 52) + .10(sin 0)=.1576N
OK.

Ax=.2231N/2.5kg= .08924
OK.
Ay=.1576N/2.5kg= .0252
Not OK. Do over.
 
Pinky did you see that the initial velocity has only a non-zero component which is its horizontal?
 
I spoke too soon before. You are a bit off track. In addition to your mistake that I already pointed out:
pinky2468 said:
tan-1(.0252/.08924)=15.8degrees(wrong answer)
It looks like you are finding the angle of the force. But you should be finding the angle of the displacement, not the force.
x=.11m/s(3.0s) + 1/2(.08924m/s^2)(3.0s)^2= wrong answer!
That's just the x-component! Don't forget to find the y-component, and then the total displacement.
 
OK! I finally got it! Thanks!
 

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