Problem 8 - Thomas Calculus

Question

Particle motion At time \(t \geq 0\) , the velocity of a body moving along the \(s\) -axis is \(v=t^{2}-4 t+3\) . a. Find the body's acceleration each time the velocity is zero. b. When is the body moving forward? Backward? c. When is the body's velocity increasing? Decreasing?

Step-by-Step Solution

Answer

a. Accelerations are -2 at \(t = 1\) and 2 at \(t = 3\). b. Moving forward for \(t < 1\) and \(t > 3\), backward for \(1 < t < 3\). c. Velocity is increasing for \(t > 2\), decreasing for \(t < 2\).

Step 1: Find Velocity Zero Points

To find when the velocity is zero, solve the equation \(v = t^2 - 4t + 3 = 0\). This is a quadratic equation, which we can factor as \((t-1)(t-3) = 0\). Thus, the velocity is zero at \(t = 1\) and \(t = 3\).

Step 2: Calculate Acceleration

The acceleration \(a(t)\) is the derivative of the velocity function \(v(t)\). So, calculate the derivative \(a(t) = \frac{d}{dt}(t^2 - 4t + 3) = 2t - 4\). Substitute \(t = 1\) to get \(a(1) = 2(1) - 4 = -2\), and \(t = 3\) to get \(a(3) = 2(3) - 4 = 2\). Thus, the body’s accelerations are -2 and 2 when the velocity is zero.

Step 3: Determine Forward and Backward Motion

The body moves forward when \(v(t) > 0\) and backward when \(v(t) < 0\). Solving \(v(t) = t^2 - 4t + 3 > 0\), we find the intervals by testing values around the roots \(t = 1\) and \(t = 3\): \(v(t)>0\) for \(t < 1\) and \(t > 3\); \(v(t)<0\) for \(1 < t < 3\). Thus, the body is moving forward when \(t < 1\) or \(t > 3\), and backward when \(1 < t < 3\).

Step 4: Determine When Velocity Is Increasing or Decreasing

The velocity is increasing when acceleration \(a(t) = 2t - 4 > 0\) and decreasing when \(a(t) < 0\). Solve \(2t - 4 > 0\) to find \(t > 2\). Thus, the velocity is increasing for \(t > 2\) and decreasing for \(t < 2\).

Key Concepts

Velocity and Acceleration

Understanding how velocity and acceleration interact is key to analyzing particle motion. Velocity describes how fast a body moves in a specific direction, while acceleration measures how quickly the velocity changes. A positive acceleration means the velocity is increasing over time, whereas a negative acceleration indicates the velocity is decreasing. To determine these values mathematically, we often use derivatives. For example, given a velocity function in terms of time, we can find acceleration by taking the derivative of that velocity function.

Imagine that velocity is like the speedometer reading on a car and acceleration is how fast you're stepping on or off the gas pedal. Both are crucial in deciding your car's movement. In the problem at hand, the body's velocity is given by the equation:

• \(v(t) = t^2 - 4t + 3\)

To compute the body’s acceleration, we derived the velocity to get:

• \(a(t) = 2t - 4\)

This acceleration tells us how the speed of the body is changing over time.

Quadratic Equations

Quadratic equations are a fundamental part of calculus and physics, often appearing in problems involving motion. A quadratic equation is a polynomial equation of the form \(ax^2 + bx + c = 0\) where \(a eq 0\). Solving these equations often brings out crucial points like maximum, minimum, or zero crossings. They are also essential for finding the time at which a particular state, such as zero velocity or turning points, occurs.

In particle motion analysis, solving a quadratic equation helps find critical moments in time. For instance, to find when the velocity becomes zero in the exercise, we solve:

• \(t^2 - 4t + 3 = 0\)

By factoring this equation, we discover:

• \((t-1)(t-3) = 0\)

This tells us that the velocity is zero at times \(t = 1\) and \(t = 3\). Using these points, we can go further to discuss changes in motion or acceleration.

Derivatives in Physics

Derivatives play a crucial role in analyzing motion in physics. They allow us to understand how a function changes over time, which is vital in understanding how particles move. A derivative is a mathematical tool that calculates the rate of change, thus showing how quickly something changes at any given point.

Take, for example, the velocity function from our exercise:

• \(v(t) = t^2 - 4t + 3\)

By deriving this function, we find the acceleration function as:

• \(a(t) = 2t - 4\)

This acceleration function shows the instantaneous rate at which the particle’s velocity is changing. A significant aspect of physics is using these derivatives to get deeper insights into how objects behave under different conditions, whether moving forwards or backwards, speeding up or slowing down.

Motion Analysis

Analyzing motion is about understanding how an object's position changes over time. It involves determining when the object moves forwards, backwards, accelerates, or decelerates. By solving mathematical equations derived from real-world scenarios, we can predict and describe motion patterns comprehensively.

In our given problem, we analyzed the motion by determining when the body's velocity is positive (moving forward) or negative (moving backward). Looking at the equation \(v(t) = t^2 - 4t + 3\), we solved it to find periods of forward (\(t < 1\) or \(t > 3\)) and backward motion (\(1 < t < 3\)). Next, we examined how acceleration influences velocity changes by solving \(2t - 4 > 0\), indicating that velocity increases for \(t > 2\) and decreases for \(t < 2\).

This comprehensive approach allows us to describe the complete behavior of a particle, predicting its future motion based on current knowledge. Understanding these principles is key in fields ranging from engineering to astrophysics.