Calculating Duck's Displacement in 2.52s with F=ma and Kinematics Equations

[scorks]

1.) A duck has a mass of 2.35 kg. As the duck paddles, a force of 0.144 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.202 N in a direction of 54.5° south of east. When these forces begin to act, the velocity of the duck is 0.130 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 2.52 s while the forces are acting.

2.) F=ma
Kinematics equations

3.) So, I worked out that the x-component of F2 is 0.117N and added that to 0.144N. I then plugged the answer (0.261N) into the F=ma formula and got a=0.11m/s^2. I then used the equation d=(Vi)t+1/2(a)(t^2) and got d=0.677m.
Then, I used the x-component of 0.261N and the Y-component of F2(the only y-component) and solved for the angle getting 31.5deg. I subtracted this from 360 to get the answer of 328.5deg relative to due east. My answers came out to be incorrect.
Please help!

 

[Delphi51]

For the east direction, I agree with your force of .261 N and a = 0.111.
I used V = Vi + at to get the final velocity of .4099 and then used
d = Vavg *t = (.13+.4099)/2*2.52 = .6803 m East.
Very close to your .677 but could that be what is getting you the wrong answer?

 

[scorks]

For the east direction, I agree with your force of .261 N and a = 0.111.
I used V = Vi + at to get the final velocity of .4099 and then used
d = Vavg *t = (.13+.4099)/2*2.52 = .6803 m East.
Very close to your .677 but could that be what is getting you the wrong answer?

Um, I think that it may be the direction that's getting me the wrong answer. What did you get for that?

 

[Delphi51]

Finished it now. We are way different on the angle. For the south distance I got about 0.2 m and an angle of less than 20 degrees south of east, which would be more than 340 degrees counterclockwise from east.

 

[Nickie]

As a scientist, it is important to approach problems with a systematic and logical approach. Let's break down the problem and go through the steps to find the displacement of the duck in 2.52 seconds using the given information and equations.

1. Identify the given information: The mass of the duck (m = 2.35 kg), the forces acting on the duck (F1 = 0.144 N, F2 = 0.202 N), the initial velocity of the duck (Vi = 0.130 m/s), and the time (t = 2.52 s).

2. Determine the acceleration: Using the formula F=ma, we can find the acceleration (a) of the duck. We know that F1 and F2 are acting on the duck in the x-direction, so we can add them together to get the total force in the x-direction (Fx = F1 + F2). Then, we can plug in the values into the formula Fx = ma and solve for a. In this case, a = 0.117 m/s^2.

3. Calculate the displacement in the x-direction: Using the kinematics equation d = Vit + 1/2at^2, we can find the displacement (dx) in the x-direction. We know that the initial velocity (Vi) is 0.130 m/s and the acceleration (a) is 0.117 m/s^2. Plugging in these values along with the time (t = 2.52 s), we get dx = 0.677 m.

4. Calculate the displacement in the y-direction: We can use the same equation (d = Vit + 1/2at^2) to find the displacement in the y-direction (dy). However, we need to consider the forces acting on the duck in the y-direction. Since F1 is in the x-direction, it does not contribute to the displacement in the y-direction. However, F2 has a y-component (Fy = F2sinθ) which is acting in the negative direction (since it is south of east). So, we can subtract this from the initial velocity in the y-direction (Vi = 0) to get the final velocity in the y-direction (Vfy = -Fy = -0.202sin54.5° = -0.157 m/s). Plugging in these values, we get dy = (-