Textbook Question
A particle moving along the x-axis has its position described by the function x = (2t3 + 2t + 1) m, where t is in s. At t = 2s, what are the particle's position?
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Ch 02: Kinematics in One Dimension
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 2, Problem 36a
The position of a particle is given by the function x = (2t3 = 6t2 + 12) m, where t is in s. At what time does the particle reach its minimum velocity? What is (vx)min?
Verified step by step guidance1
Step 1: Start by identifying the velocity function from the given position function x(t). Velocity is the derivative of position with respect to time. Compute v(t) = dx/dt. For x(t) = 2t^3 - 6t^2 + 12, differentiate term by term to get v(t) = 6t^2 - 12t.
Step 2: To find the time at which the particle reaches its minimum velocity, determine the critical points of v(t). This involves finding the derivative of v(t), which is the acceleration a(t). Compute a(t) = dv/dt = d(6t^2 - 12t)/dt = 12t - 12.
Step 3: Set the acceleration a(t) equal to zero to find the critical points. Solve 12t - 12 = 0 for t. This will give the time at which the velocity is at a minimum or maximum.
Step 4: Verify whether the critical point corresponds to a minimum velocity by analyzing the second derivative of v(t) or by checking the behavior of v(t) around the critical point. The second derivative of v(t) is d^2v/dt^2 = 12, which is positive, indicating a minimum velocity at the critical point.
Step 5: Substitute the critical time value back into the velocity function v(t) = 6t^2 - 12t to calculate the minimum velocity (vx)min. This gives the value of the minimum velocity at the determined time.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Position Function
The position function describes the location of a particle as a function of time. In this case, the position is given by x(t) = 2t^3 - 6t^2 + 12, which is a polynomial function. Understanding this function is crucial for determining the particle's motion and calculating its velocity by taking the derivative with respect to time.
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Velocity
Velocity is the rate of change of position with respect to time, mathematically represented as the derivative of the position function. For the given position function, the velocity function v(t) can be found by differentiating x(t). The minimum velocity occurs when the derivative of the velocity function equals zero, indicating a change in direction or speed.
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Critical Points
Critical points are values of the independent variable (in this case, time) where the derivative of a function is zero or undefined. These points are essential for finding local extrema, such as minimum or maximum values. To find the minimum velocity of the particle, one must identify the critical points of the velocity function derived from the position function.
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Related Practice
Textbook Question
A block is suspended from a spring, pulled down, and released. The block's position-versus-time graph is shown in FIGURE P2.38. At what times is the velocity zero? At what times is the velocity most positive? Most negative?
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Textbook Question
The vertical position of a particle is given by the function y = (t2 - 4t + 2) m, where t is in s. What is the particle's position at that time?
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Textbook Question
A particle moving along the x-axis has its veocity described by the function vx = 2t2 m/s, where t is in s. itsinitial position is x0 = 1 m at t0 = 0 s. At t = 1 s, what are the particle's position?
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Textbook Question
A block is suspended from a spring, pulled down, and released. The block's position-versus-time graph is shown in FIGURE P2.38. Draw a reasonable velocity-versus-time graph.
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Textbook Question
The position of a particle is given by the function x = (2t3 - 6t2 + 12) m, where t is in s. At what time is the acceleration zero?
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