A rocket starts from rest and moves upward from the surface of the earth. For the first 10.010.010.0 s of its motion, the vertical acceleration of the rocket is given by ay=(2.80a_{y}=(2.80ay=(2.80 m/s3)t)t)t, where the +y+y+y-direction is upward. What is the speed of the rocket when it is 325325 m above the surface of the earth?

Ch 02: Motion Along a Straight Line

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

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Young & Freedman Calc 15th Edition

Ch 02: Motion Along a Straight Line

Problem 49b

Chapter 2, Problem 49b

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 $a_{y}=(2.80$ m/s³ $)t$ , where the $+y$ -direction is upward. What is the speed of the rocket when it is $325$ m above the surface of the earth?

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Identify the given information: The rocket starts from rest, so the initial velocity (v0) is 0 m/s. The acceleration is given as a function of time: ay = (2.80 m/s^3)t. We need to find the speed of the rocket when it is 325 m above the surface.

Use the kinematic equation for position with variable acceleration: s = ∫v dt, where v = ∫a dt. First, find the velocity function by integrating the acceleration: v(t) = ∫(2.80 m/s^3)t dt.

Calculate the velocity function: v(t) = (2.80 m/s^3)(t^2)/2 + C. Since the rocket starts from rest, the constant C is 0. Therefore, v(t) = 1.40 m/s^3 * t^2.

Use the position function: s(t) = ∫v(t) dt = ∫(1.40 m/s^3 * t^2) dt. Integrate to find the position function: s(t) = (1.40 m/s^3)(t^3)/3 + C. Since the initial position is 0, C is 0, so s(t) = 0.467 m/s^3 * t^3.

Solve for the time when the rocket is 325 m above the surface: Set s(t) = 325 m and solve for t. Once t is found, substitute it back into the velocity function v(t) = 1.40 m/s^3 * t^2 to find the speed of the rocket at that height.

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Key Concepts

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

Kinematics Equations

Kinematics equations describe the motion of objects without considering the forces that cause the motion. They relate displacement, velocity, acceleration, and time. In this problem, the kinematic equation v = u + at is crucial, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Variable Acceleration

Variable acceleration occurs when the acceleration of an object changes with time. In this scenario, the rocket's acceleration is given as a function of time, ay = (2.80 m/s³)t. To find the velocity, we need to integrate this acceleration function with respect to time, considering the initial conditions.

Integration in Physics

Integration is a mathematical process used to find quantities like displacement or velocity when given a rate of change, such as acceleration. For this problem, integrating the acceleration function ay = (2.80 m/s³)t with respect to time will yield the velocity function, which can then be evaluated to find the speed at a specific height.

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Textbook Question

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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