A duck has a mass of 2.8 kg. As the duck paddles, a force of 0.09 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 47° 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 2.5 s while the forces are acting
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To find the displacement of the duck during the 2.5-second interval, we need to consider the net force acting on it and how it affects the duck's velocity.
First, let's break down the forces acting on the duck:
1. Force exerted due east: 0.09 N
2. Current force: 0.20 N, at a direction 47° south of east
The net force acting on the duck is the vector sum of these two forces. We can calculate the horizontal and vertical components separately.
Horizontal component:
The force exerted due east is already in the horizontal direction, so its horizontal component is 0.09 N.
Vertical component:
To find the vertical component of the current force, we need to calculate the sine of the angle: sin(47°) = 0.7314.
So, the vertical component of the current force is 0.7314 * 0.20 N = 0.14628 N.
Now, let's find the net force in the horizontal and vertical directions:
Horizontal net force = force exerted due east = 0.09 N
Vertical net force = vertical component of current force = 0.14628 N
Since there is no horizontal net force, the duck's velocity in the horizontal direction remains constant at 0.12 m/s.
In the vertical direction, the net force will cause a change in velocity. We can use Newton's second law (F = ma) to find the acceleration in the vertical direction:
Vertical net force = mass * acceleration
0.14628 N = 2.8 kg * acceleration
Solving for acceleration:
acceleration = 0.14628 N / 2.8 kg ≈ 0.0522 m/s²
Now, we can calculate the change in velocity in the vertical direction during the 2.5-second interval:
Δvelocity = acceleration * Δtime
Δvelocity = 0.0522 m/s² * 2.5 s ≈ 0.1305 m/s
Since the initial velocity in the vertical direction is 0 (the duck is not moving up or down), the final velocity is 0.1305 m/s downward.
Now, we can find the vertical displacement using the formula:
Vertical displacement = (initial velocity + final velocity) / 2 * Δtime
Vertical displacement = (0 + (-0.1305 m/s)) / 2 * 2.5 s ≈ -0.16313 m
The negative sign indicates that the displacement is downward.
The horizontal displacement remains the same since there is no net force in the horizontal direction.
Therefore, the magnitude of the displacement that the duck undergoes in 2.5 s is the square root of the sum of the squared horizontal and vertical displacements:
Magnitude of displacement = √((horizontal displacement)² + (vertical displacement)²)
Magnitude of displacement = √((0 m)² + (-0.16313 m)²) ≈ 0.16313 m
The direction of the displacement is given by the angle:
Direction = arctan(vertical displacement / horizontal displacement)
Direction = arctan(-0.16313 m / 0) = -90° (relative to due east)
So, the magnitude of the displacement is approximately 0.16313 m, and the direction is 90° south of east.