/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A particle moves along the \(x\)... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle moves along the \(x\) axis according to the equation \(x=2.00+3.00 t-1.00 t^{2},\) where \(x\) is in meters and \(t\) is in seconds. At \(t=3.00 \mathrm{s},\) find (a) the position of the particle, (b) its velocity, and (c) its acceleration.

Short Answer

Expert verified
(a) The position of the particle at \(t=3.00s\) is 2.00m. (b) Its velocity at \(t=3.00s\) is -3.00m/s. (c) Its acceleration at \(t=3.00s\) is -2.00m/s^2.

Step by step solution

01

Find the position of the particle at \(t=3.00s\)

To find the position of the particle at the given time, substitute \(t=3.00s\) in the position function \(x(t) = 2.00 + 3.00t -1.00t^2\). This gives: \(x(3) = 2.00 + 3.00*3 -1.00*(3^2)= 2.00 + 9.00 - 9.00 = 2.00m\).
02

Find the velocity of the particle at \(t=3.00s\)

The velocity function is the first derivative of the position function, hence, \(v(t) = dx/dt = 3.00 - 2.00t\). Substitute \(t=3.00s\) into the velocity function to obtain \(v(3) = 3.00 - 2.00*3 = -3.00m/s\).
03

Find the acceleration of the particle at \(t=3.00s\)

The acceleration function is the second derivative of the position function, \(a(t) = dv/dt = -2.00m/s^2\). Since this function is independent of \(t\), the particle's acceleration remains constant at -2.00m/s^2 at any time, including at \(t=3.00s\).

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Key Concepts

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

Particle Motion in One Dimension
Understanding particle motion in one dimension is pivotal for grasping the fundamentals of kinematics—the branch of physics that describes the motion of objects. When we consider the movement of a particle along a straight line, such as the x-axis, we are examining particle motion in one dimension. This scenario simplifies the complex real-world movements into a format that’s easier to analyze mathematically.

For instance, the particle described in our exercise has its motion defined by the equation \(x(t) = 2.00 + 3.00t - 1.00t^2\), where \(x\) represents the position of the particle in meters at any time \(t\), measured in seconds. When we want to determine the particle's position at a specific time—say, \(t = 3.00s\)—we substitute this value into our position function to calculate the precise location on the x-axis.

Understanding this concept is essential for predicting where a particle will be at a certain point in time, which has practical applications in areas ranging from physics experiments to vehicle navigation systems.
Velocity and Acceleration
Velocity and acceleration are fundamental concepts that describe how the speed and direction of a particle’s motion change over time. To find the velocity of a particle, we look at the rate at which its position changes, which is calculated as the first derivative of the position function with respect to time. In the context of our exercise, the velocity function derived from \(x(t) = 2.00 + 3.00t - 1.00t^2\) is \(v(t) = dx/dt = 3.00 - 2.00t\).

Acceleration, on the other hand, measures how quickly the velocity changes and is determined by taking the second derivative of the position function, or equivalently, the first derivative of the velocity function with respect to time. For the particle we’re studying, the acceleration is a constant \(-2.00m/s^2\), indicating that its velocity decreases by this amount every second.

Interpreting Negative Acceleration

When the acceleration is negative, as it is in our exercise, it means the particle is slowing down in the direction of motion or speeding up in the opposite direction—often referred to as deceleration.
Derivatives in Physics
Derivatives play a crucial role in physics by providing a mathematical tool for modeling rates of change. These rates of change are at the heart of understanding motion, forces, energy, and other key physical concepts. In kinematics, especially, derivatives allow us to calculate an object's velocity and acceleration based on its position function.

In our exercise scenario, the derivatives of the position function \(x(t) = 2.00 + 3.00t - 1.00t^2\) enable us to describe the entire state of motion. The first derivative gives us the velocity, representing the rate of change of position with time, and the second derivative gives us the acceleration, representing the rate of change of velocity with time. It’s these successive derivatives that illuminate the dynamic qualities of particle motion, which would be impossible to capture by position alone.

Practical Implications

Understanding how to apply derivatives in physics can significantly enhance a student's capability to predict and analyze the behavior of moving objects, making it an indispensable part of their education in science, technology, engineering, and mathematics (STEM) fields.

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Most popular questions from this chapter

An object moving with uniform acceleration has a velocity of \(12.0 \mathrm{cm} / \mathrm{s}\) in the positive \(x\) direction when its \(x\) coordinate is \(3.00 \mathrm{cm} .\) If its \(x\) coordinate \(2.00 \mathrm{s}\) later is \(-5.00 \mathrm{cm},\) what is its acceleration?

A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of \(80.0 \mathrm{m} / \mathrm{s}\) at ground level. Its engines then fire and it accelerates upward at \(4.00 \mathrm{m} / \mathrm{s}^{2}\) until it reaches an altitude of \(1000 \mathrm{m} .\) At that point its engines fail and the rocket goes into free fall, with an acceleration of \(-9.80 \mathrm{m} / \mathrm{s}^{2} .\) (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? (You will need to consider the motion while the engine is operating separate from the free-fall motion.)

A truck covers \(40.0 \mathrm{m}\) in \(8.50 \mathrm{s}\) while smoothly slowing down to a final speed of \(2.80 \mathrm{m} / \mathrm{s} .\) (a) Find its original speed. (b) Find its acceleration.

In a \(100-\mathrm{m}\) linear accelerator, an electron is accelerated to \(1.00 \%\) of the speed of light in \(40.0 \mathrm{m}\) before it coasts for \(60.0 \mathrm{m}\) to a target. (a) What is the electron's acceleration during the first \(40.0 \mathrm{m} ?\) (b) How long does the total flight take?

A car has an initial velocity \(v_{0}\) when the driver sees an obstacle in the road in front of him. His reaction time is \(\Delta t_{r}\) and the braking acceleration of the car is \(a\). Show that the total stopping distance is $$s_{\text {stop }}=v_{0} \Delta t_{r}-v_{0}^{2} / 2 a$$,Remember that \(a\) is a negative number.

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