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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
Chapter 4, Problem 9a

A particle's trajectory is described by x=(12t32t2)mandy=(12t22t)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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Step 1: Understand the problem. The particle's trajectory is given by two equations: x = (1/2t^2 - 2t^2) m and y = (1/2t^2 - 2t) m. We need to find the particle's position (x, y) and speed at two specific times: t = 0 s and t = 4 s.
Step 2: Calculate the position at t = 0 s. Substitute t = 0 into the equations for x and y. For x, substitute t = 0 into x = (1/2t^2 - 2t^2). For y, substitute t = 0 into y = (1/2t^2 - 2t). This will give the coordinates (x, y) at t = 0 s.
Step 3: Calculate the position at t = 4 s. Substitute t = 4 into the equations for x and y. For x, substitute t = 4 into x = (1/2t^2 - 2t^2). For y, substitute t = 4 into y = (1/2t^2 - 2t). This will give the coordinates (x, y) at t = 4 s.
Step 4: Determine the speed at t = 0 s and t = 4 s. To find the speed, calculate the magnitude of the velocity vector. First, find the velocity components by differentiating x and y with respect to t. For x, differentiate x = (1/2t^2 - 2t^2) to get v_x. For y, differentiate y = (1/2t^2 - 2t) to get v_y. Then, calculate the speed using the formula: speed = sqrt(v_x^2 + v_y^2).
Step 5: Evaluate the speed at t = 0 s and t = 4 s. Substitute t = 0 and t = 4 into the expressions for v_x and v_y obtained in Step 4. Use these values to calculate the speed at each time using the formula: speed = sqrt(v_x^2 + v_y^2).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinematics

Kinematics is the branch of physics that describes the motion of objects without considering the forces that cause the motion. It involves concepts such as position, velocity, and acceleration, which are essential for analyzing the trajectory of a particle. In this question, understanding kinematics allows us to determine the particle's position and speed at specific time intervals.
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Position and Displacement

Position refers to the location of a particle in space at a given time, while displacement is the change in position from one point in time to another. The equations provided describe the particle's position in two dimensions (x and y) as functions of time. Evaluating these equations at specific times (t = 0 s and t = 4 s) will yield the particle's exact position in the coordinate system.
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Velocity

Velocity is a vector quantity that describes the rate of change of position with respect to time. It is calculated as the derivative of the position function with respect to time. In this problem, finding the velocity at the specified times involves differentiating the position equations to determine how fast and in what direction the particle is moving at those moments.
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Related Practice
Textbook Question

You have a remote-controlled car that has been programmed to have velocity v=(3ti+2t2j)m/s\(\mathbf{v}\) = (-3t\(\mathbf{i}\) + 2t^2\(\mathbf{j}\)) \, \(\text{m/s}\), where t is in s. At t = 0 s, the car is at r0=(3.0i+2.0j)m\(\mathbf{r}\)_0 = (3.0\(\mathbf{i}\) + 2.0\(\mathbf{j}\)) \, \(\text{m}\). What are the car's position vector?

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Textbook Question

A particle's trajectory is described by x=(12t32t2)mandy=(12t22t)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 is the particle's direction of motion, measured as an angle from the xx-axis, at t=0 st=0\(\text{ s}\) and t=4 st=4\(\text{ s}\)?

Textbook Question

A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 100 m above the glacier at a speed of 150 m/s. How far short of the target should it drop the package?

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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

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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