Newtons Law Problem involving kinematic equation

[lolcheelol]

Homework Statement

A duck has a mass of 3.0 kg. As the duck paddles, a force of 0.06 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.13 N in a direction of 60° south of east. When these forces begin to act, the velocity of the duck is 0.12 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 2.0 s while the forces are acting.

Homework Equations

ƩF = m*a
D = VoT + (1/2)aT^2

The Attempt at a Solution

Fx = 0.06 + cos(60)*.13 = .125 m/s
Fy = sin(60)*.13 = .113 m/s

sqrt(.125^2 + .113^2) = .1685N

.1685/3.0 = 0.0561 m/s^2

D = (0.12)(2) + (1/2)(.0561)(2)^2 = .3522
This is where I am confused, as I think I thought I was doing everything correct up until this point. When I enter this into the answer sheet it says it is wrong. I mistakenly used .13 for Fy instead of .113 and it gave me .326 m which is the correct answer.

Also I know to find the direction you use arctan(Fy/Fx) but I can't find it...

Any help is greatly appreciated.

Thank you.

 

[gneill]

Homework Statement

A duck has a mass of 3.0 kg. As the duck paddles, a force of 0.06 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.13 N in a direction of 60° south of east. When these forces begin to act, the velocity of the duck is 0.12 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 2.0 s while the forces are acting.

Homework Equations

ƩF = m*a
D = VoT + (1/2)aT^2

The Attempt at a Solution

Fx = 0.06 + cos(60)*.13 = .125 m/s
Fy = sin(60)*.13 = .113 m/s

Careful with the units! Those are forces, not velocities.

sqrt(.125^2 + .113^2) = .1685N

That's better! That's the magnitude of the net force.

.1685/3.0 = 0.0561 m/s^2

Okay, that's the magnitude of the acceleration. But what are you going to do with that?

D = (0.12)(2) + (1/2)(.0561)(2)^2 = .3522

Uh oh

The acceleration vector and initial velocity vector do not lie in the same direction. So you can't just use their magnitudes to find the displacement.

Instead, keep the velocity and acceleration broken into components and perform the displacement calculation separately on each; Find Δx and Δy, then combine the results into a single displacement. These components will also give you the direction of the displacement.

 

[lolcheelol]

So do you mean

x = (0.12)(2)+(1/2)(.125)(2)^2 = .49
y = (0.12)(2)+(1/2)(.113)(2)^2 = .466

x+y = .49 + .466 = .956

then do arctan(.466/.49)

 

[gneill]

Almost. The initial velocity will also have separate components (and they're not the same). What are they?
Also, be sure that you have accelerations to plug into the equation (not forces).
Also, beware of the signs on your components! Draw a diagram: the current is applying its force in a direction 60° SOUTH of east.

 

[lolcheelol]

The initial velocity for x would be 0.12 and for y=0.

x acceleration would be .125/3.0 because of F/M = A
y acceleration would be .113/3.0 " "

x = .041 m/s^2
y = .037 m/s^2

x = (0.12)(2)+(1/2)(.041)(2)^2 = .322
y = (0)(2)+(1/2)(.037)(2)^2 = .074

then would you do sqrt[.322^2 + .074^2] = .330
or
.322 + .074 = .396

 

[gneill]

The initial velocity for x would be 0.12 and for y=0.

x acceleration would be .125/3.0 because of F/M = A
y acceleration would be .113/3.0 " "

Did you check the signs for the components? What does an angle "South of East" imply?

Ax = .041 m/s^2
Ay = .037 m/s^2

x = (0.12)(2)+(1/2)(.041)(2)^2 = .322
y = (0)(2)+(1/2)(.037)(2)^2 = .074

Other than the component sign issue, you might want to keep a few more decimal places for intermediate results. I think that truncation and rounding errors are creeping into your results.

then would you do sqrt[.322^2 + .074^2] = .330
or
.322 + .074 = .396

Definitely the square root of the sum of the squares. The components of a Cartesian vector always "add in quadrature" that way.

 

[lolcheelol]

Thanks for being so patient with me and all your help. I finally got it. Keeping separate the velocitys and acceleration components really helped.