A rocket starts from rest and moves upward from the surface of the earth. For the first 10.010.0 s of its motion, the vertical acceleration of the rocket is given by ay=(2.80a_{y}=(2.80 m/s3)t)t, where the +y+y-direction is upward. What is the height of the rocket above the surface of the earth at t=10.0t = 10.0 s?

Ch 02: Motion Along a Straight Line

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 02: Motion Along a Straight Line

Problem 49a

Chapter 2, Problem 49a

A rocket starts from rest and moves upward from the surface of the earth. For the first $10.0$ s of its motion, the vertical acceleration of the rocket is given by $a sub y equals open paren 2.80$ m/s³ $) t$ , where the $+ y$ -direction is upward. What is the height of the rocket above the surface of the earth at $t equals 10.0$ s?

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Identify the given information: The rocket starts from rest, meaning its initial velocity is 0 m/s. The vertical acceleration is given by $ a_y = 2.80 \text{ m/s}^3 \cdot t $, and we need to find the height at $ t = 10.0 \text{ s} $.

Understand the relationship between acceleration, velocity, and displacement. Since acceleration is a function of time, integrate the acceleration function $ a_y(t) = 2.80 \text{ m/s}^3 \cdot t $ with respect to time to find the velocity function $ v_y(t) $.

Perform the integration: $ v_y(t) = \int a_y(t) \, dt = \int 2.80 \text{ m/s}^3 \cdot t \, dt = 1.40 \text{ m/s}^3 \cdot t^2 + C $. Since the initial velocity is 0 m/s, $ C = 0 $, so $ v_y(t) = 1.40 \text{ m/s}^3 \cdot t^2 $.

Integrate the velocity function $ v_y(t) = 1.40 \text{ m/s}^3 \cdot t^2 $ to find the displacement function $ y(t) $. $ y(t) = \int v_y(t) \, dt = \int 1.40 \text{ m/s}^3 \cdot t^2 \, dt = \frac{1.40}{3} \text{ m/s}^3 \cdot t^3 + C $. Since the initial position is 0, $ C = 0 $, so $ y(t) = \frac{1.40}{3} \text{ m/s}^3 \cdot t^3 $.

Substitute $ t = 10.0 \text{ s} $ into the displacement function $ y(t) = \frac{1.40}{3} \text{ m/s}^3 \cdot t^3 $ to find the height of the rocket above the surface of the earth at $ t = 10.0 \text{ s} $.

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Key Concepts

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

Kinematics in One Dimension

Kinematics involves the study of motion without considering its causes. In one-dimensional motion, key equations relate displacement, velocity, acceleration, and time. For this problem, understanding how acceleration affects velocity and displacement over time is crucial to determine the rocket's height.

Integration in Physics

Integration is a mathematical tool used to find quantities like displacement from acceleration. Since acceleration is given as a function of time, integrating this function over the specified time interval provides the velocity, and further integration gives the displacement, which is the rocket's height at t = 10.0 s.

Variable Acceleration

Variable acceleration means acceleration changes with time, unlike constant acceleration scenarios. Here, the acceleration is a function of time, ay = (2.80 m/s³)t, requiring calculus to find velocity and displacement. Understanding how to handle variable acceleration is essential for solving the problem accurately.

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