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True or False The acceleration of a particle is the second derivative of the position function. Justify your answer.

Short Answer

Expert verified
Yes, it is true that the acceleration of a particle is the second derivative of the position function. This is because acceleration is the rate change of velocity, which is already the rate of change of the position.

Step by step solution

01

Physics Definitions

First, it's fundamental to understand the definitions of position, velocity, and acceleration. The position of an object refers to its location at a certain point of time. Velocity describes the rate of change of the position over time, it essentially is the speed and direction. Acceleration, on the other hand, is the rate of change of velocity over time.
02

Mathematical Relation

In calculus, the rate of change of a function is represented by its derivative. Therefore, the velocity (which is the rate of change of position) will be the first derivative of the position function, and acceleration (which is the rate of change of velocity) will be the derivative of the velocity function.
03

Final Explanation

By connecting those concepts, we can assert that the acceleration, being the rate of change of velocity (which is already the rate of change of position), is indeed the second derivative of the position function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives
Derivatives in calculus are like magical tools. They help us understand how fast something is changing. When we talk about functions, a derivative can tell us the rate at which one quantity changes with respect to another. For instance, if you have a function that describes your position over time, its derivative will tell you how your position's changing – essentially giving you your velocity.
  • A derivative is found using calculus techniques like differentiation.
  • It tells us the slope of the function at any given point, which represents the rate of change.
Think of it as taking a snapshot of how steep the line of a graph is at any point. A straight, horizontal line means zero change, while a steep line shows fast change. This concept underpins many applications of calculus, especially in physical sciences, where it aids in tracking motion, growth, and decay.
Position Function
The position function is simply the function that describes where an object is located at any given time. Imagine it as keeping track of your location on a map as time passes. Generally, the position function is denoted by \(s(t)\), where \(t\) stands for time.
  • This function tells you your exact location at each time point.
  • From this function, derivatives can help us find other key metrics like velocity and acceleration.
For example, if you have a position function, \(s(t) = 5t^2\), this might describe a situation where your distance from the starting point grows as a square of time. Using differentiation, the position function makes it easier to calculate how quickly you're moving (velocity) and how that velocity changes (acceleration). Understanding the position function is crucial because it connects directly to where an object is and how it's moving.
Acceleration
Acceleration is like the speedometer of your velocity. It tells us how quickly the velocity itself is changing. In terms of calculus, acceleration is the second derivative of the position function.
  • First, we derive the position function to get velocity.
  • Then, we derive the velocity function to find the acceleration.
This means you're essentially looking at changes of changes – acceleration describes how quickly things are speeding up or slowing down. For example, if you're in a car that's speeding up, the rate at which your speed increases is your acceleration. It's a crucial concept because it explains not just the movement, but how motion itself is changing over time. This makes it vital in physics and engineering for understanding dynamics and movement forces. So, indeed, when you derive twice from the position function, you arrive at acceleration, confirming the statement in our exercise as "True."

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Most popular questions from this chapter

Particle Motion A particle moves along a line so that its position at any time \(t \geq 0\) is given by the function \(s(t)=\) \(-t^{3}+7 t^{2}-14 t+8\) where \(s\) is measured in meters and \(t\) is measured in seconds. (a) Find the instantaneous velocity at any time t. (b) Find the acceleration of the particle at any time t. (c) When is the particle at rest? (d) Describe the motion of the particle. At what values of t does the particle change directions?

Even and Odd Functions (a) Show that if \(f\) is a differentiable even function, then \(f^{\prime}\) is an odd function. (b) Show that if \(f\) is a differentiable odd function, then \(f^{\prime}\) is an even function.

In Exercises 11 and \(12,\) a weight hanging from a spring see Figure 3.38 ) bobs up and down with position function \(s=f(t)\) in meters, \(t\) in seconds. What are its velocity and acceleration at time \(t ?\) Describe its motion. (a) Find the body's velocity, speed, and acceleration at time \(t .\) (b) Find the body's velocity, speed, and acceleration at time \(t=\pi / 4 .\) (c) Describe the motion of the body. $$s=7 \cos t$$

Find points on the curve \(x^{2}+x y+y^{2}=7\) (a) where the tangent is parallel to the \(x\) -axis and (b) where the tangent is parallel to the \(y\) -axis. (In the latter case, \(d y / d x\) is not defined, but \(d x / d y\) is. What value does \(d x / d y\) have at these points?

The Cissoid of Diocles (dates from about 200 B.c.) (a) Find equations for the tangent and normal to the cissoid of Diocles, $$y^{2}(2-x)=x^{3}$$ at the point (1, 1) as pictured below. (b) Explain how to reproduce the graph on a grapher.

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