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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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 36b
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?
Verified step by step guidance1
Step 1: Start by recalling that acceleration is the second derivative of the position function with respect to time. The position function is given as x(t) = 2t^3 - 6t^2 + 12.
Step 2: Compute the first derivative of the position function to find the velocity function. Use the power rule for differentiation: v(t) = dx/dt = d/dt(2t^3 - 6t^2 + 12).
Step 3: Differentiate the velocity function to find the acceleration function. Again, use the power rule: a(t) = dv/dt = d/dt(6t^2 - 12t).
Step 4: Set the acceleration function equal to zero to find the time when the acceleration is zero. Solve the equation a(t) = 0 for t.
Step 5: Solve the resulting equation for t, ensuring you check for all possible solutions. These values of t represent the times when the acceleration is zero.
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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 x is given by the polynomial equation x = 2t^3 - 6t^2 + 12, which indicates how the particle's position changes over time. Understanding this function is crucial for determining the particle's velocity and acceleration.
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Intro to Wave Functions
Velocity
Velocity is the rate of change of position with respect to time, mathematically represented as the first derivative of the position function. For the given position function, the velocity v can be found by differentiating x with respect to t. This step is essential for analyzing the motion of the particle and determining when the acceleration is zero.
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Acceleration
Acceleration is the rate of change of velocity with respect to time, represented as the second derivative of the position function. To find when the acceleration is zero, one must differentiate the velocity function obtained from the position function and set it equal to zero. This indicates points in time where the particle is not changing its velocity, which is critical for understanding its motion.
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Related Practice
Textbook Question
A particle's velocity is described by the function vₓ = (t² - 7t + 10) m/s, where t is in s. What is the particle's acceleration at each of the turning points?
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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
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?
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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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