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Ch 04: Kinematics in Two Dimensions
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 04: Kinematics in Two DimensionsProblem 6b
Chapter 4, Problem 6b

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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Step 1: Analyze the graph for the x-component of velocity (Vx). The graph shows that Vx is constant at 3 cm/s over the time interval from t = 0 to t = 5 seconds. This indicates uniform motion in the x-direction.
Step 2: Use the formula for displacement in the x-direction under constant velocity: x = Vx * t. Here, Vx = 3 cm/s and t = 5 s.
Step 3: Analyze the graph for the y-component of velocity (Vy). Since the graph for Vy is not provided, we assume Vy = 0 cm/s, meaning there is no motion in the y-direction.
Step 4: Calculate the total displacement. Since there is no motion in the y-direction, the displacement in the y-direction is zero. The total displacement is therefore equal to the displacement in the x-direction.
Step 5: Combine the displacements in the x and y directions to find the distance from the origin. Use the Pythagorean theorem: distance = sqrt(x^2 + y^2). Substitute x and y values to find the distance.

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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, typically in the x and y directions. In this scenario, the x-component of the velocity is constant, as indicated by the horizontal line in the graph. Understanding these components is crucial for determining the puck's position over time.
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Kinematics Equations

Kinematics equations describe the motion of objects under constant velocity or acceleration. For an object moving with constant velocity, the position can be calculated using the formula: position = initial position + velocity × time. This principle is essential for finding the puck's position at a given time.
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Frictionless Motion

In a frictionless environment, an object continues to move at a constant velocity unless acted upon by an external force. This simplifies the analysis of motion, as there are no opposing forces to consider. The hockey puck's motion on a frictionless table allows us to apply straightforward kinematic principles without complicating factors.
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Related Practice
Textbook Question

A particle's trajectory is described by x=(12t3−2t2)mandy=(12t2−2t)m,x = \(\left\)(\(\frac{1}{2}\) t^3 - 2t^2\(\right\)) \, \(\text{m}\) \(\quad\) \(\text{and}\) \(\quad\) y = \(\left\)(\(\frac{1}{2}\) t^2 - 2t\(\right\)) \, \(\text{m}\), where tt is in ss. What are the particle's position and speed at t=0 st=0\(\text{ s}\) and t=4 st=4\(\text{ s}\)?

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Textbook Question

At this instant, the particle is speeding up and curving upward. What is the direction of its acceleration?

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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 EX4.6 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.

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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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