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

Problema 49a

Capitolo 2, Problema 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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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.

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08:29

Kinematics in 2D

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.

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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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Impulse with Variable Forces

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