Textbook Question
At this instant, the particle is speeding up and curving upward. What is the direction of its acceleration?
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Ch 04: Kinematics in Two Dimensions
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 4, Problem 6a
A rocket-powered hockey puck moves on a horizontal frictionless table. FIGURE EX4.6 shows graphs of vx and vy, the x- and y-components of the puck's velocity. The puck starts at the origin. In which direction is the puck moving at t = 2s? Give your answer as an angle from the x-axis.
Verified step by step guidance1
Step 1: Analyze the graph for the x-component of velocity (v_x). From the graph, we observe that v_x is constant at -3 m/s for all times, including t = 2s. This indicates that the puck is moving in the negative x-direction with a constant velocity.
Step 2: Analyze the graph for the y-component of velocity (v_y). At t = 2s, the graph shows that v_y is decreasing linearly. By reading the graph, we find that v_y at t = 2s is approximately 3 m/s. This indicates that the puck is moving in the positive y-direction at this instant.
Step 3: Combine the x- and y-components of velocity to determine the direction of motion. The puck's velocity vector at t = 2s can be represented as v = (-3 m/s) î + (3 m/s) ĵ, where î and ĵ are unit vectors in the x- and y-directions, respectively.
Step 4: Calculate the angle of the velocity vector relative to the x-axis. The angle θ can be found using the formula tan(θ) = v_y / v_x. Substituting the values, tan(θ) = 3 / -3.
Step 5: Solve for θ using the inverse tangent function. Since v_x is negative and v_y is positive, the angle will be in the second quadrant. Use θ = arctan(v_y / v_x) and adjust for the quadrant to find the direction of motion relative to the x-axis.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Velocity Components
Velocity can be broken down into its components along the x and y axes. The x-component (vx) indicates motion in the horizontal direction, while the y-component (vy) indicates motion in the vertical direction. Understanding these components is crucial for determining the overall direction of an object's motion, especially when analyzing two-dimensional motion.
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Calculating Velocity Components
Graph Interpretation
Graphs of velocity versus time provide visual representations of how an object's velocity changes over time. In this case, the x-component graph shows a constant velocity of -3 m/s, indicating motion to the left, while the y-component graph shows a linear decrease from 5 m/s to 0 m/s, indicating a change in vertical motion. Analyzing these graphs helps in determining the object's trajectory at specific time intervals.
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Direction of Motion
The direction of motion can be expressed as an angle relative to a reference axis, typically the x-axis. This angle can be calculated using the arctangent function, which relates the y-component to the x-component of velocity. By determining the signs and magnitudes of these components at a given time, one can find the angle of the puck's motion, which is essential for understanding its trajectory.
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Related Practice
Textbook Question
Complete the motion diagram by adding acceleration vectors.
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Textbook Question
A rocket-powered hockey puck moves on a horizontal frictionless table. FIGURE EX4.6 shows graphs of vx and vy, the x- and y-components of the puck's velocity. The puck starts at the origin. How far from the origin is the puck at t = 5s?
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Textbook Question
A particle moving in the xy-plane has velocity v = (2ti + (3-t2)j) m/s, where t is in s. What is the particle's acceleration vector at t = 4s?
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Textbook Question
Is this particle curving upward, curving downward, or moving in a straight line?
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Textbook Question
A rocket-powered hockey puck moves on a horizontal frictionless table. Figure EX4.7 shows graphs of vx and vy the x- and y-components of the puck's velocity. The puck starts at the origin. What is the magnitude of the puck's acceleration at t = 5s?
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