Question

Particle Motion The position \((x-\) coordinate) of a particle moving on the line \(y=2\) is given by \(x(t)=2 t^{3}-13 t^{2}+22 t-5\) where is time in seconds. (a) Describe the motion of the particle for \(t \geq 0\) . (b) When does the particle speed up? slow down? (c) When does the particle change direction? (d) When is the particle at rest? (e) Describe the velocity and speed of the particle. (f) When is the particle at the point \((5,2) ?\)

Short answer

The particle moves right for \(0 <= t < 2\) and \(11/3 < t\), and to the left for \(2 < t < 11/3\). It changes direction at \(t = 2\) and \(t = 11/3\). The particle speeds up for \(2 < t < 11/3\) and \(11/3 < t\), and slows down for \(0 < t < 2\). The particle is at rest at \(t = 2\) and \(t = 11/3\). The velocity and speed of the particle are described by \(v(t) = 6t^{2} - 26t + 22\) and |v(t)| respectively. The particle is at position \((5,2)\) at \(t≈1.35\) s.

Step-by-step solution

- Dissect the Given Expression

Identify the position function \(x(t) = 2t^{3} - 13t^{2} + 22t - 5\). This implies that the particle's position at any given time \(t\) is determined by this function.

- Find the Velocity Function

Compute the first derivative of \(x(t)\) to get the velocity function \(v(t)\). Using the power rule, take the derivative of each term one at a time. The derivative of position function \(x(t)\) gives us the velocity function \(v(t) = 6t^{2} - 26t + 22\).

- Find the Acceleration Function

Compute the second derivative of \(x(t)\) to get the acceleration function \(a(t)\). The derivative of velocity function \(v(t)\) gives us the acceleration function \(a(t) = 12t - 26\).

- Find the Time for Velocity Zero

Find the zeros of the velocity equation \(v(t) = 0\), it results \(t = 2\), and \(t = 11/3\). These time points are crucial as the particle is at rest at these moments. The direction of particles' movement changes at these moments.

- Describe Motion Behaviour

Using the roots from Step 4, we can establish when the particle moves to the left or to the right. When \(0 <= t < 2\) and \(11/3 < t\), the velocity is positive so the particle moves to the right. When \(2 < t < 11/3\), the velocity is negative so the particle moves to the left.

- Determine Direction Change

The particle changes direction when the velocity changes sign which occurs at \(t = 2\) and \(t = 11/3\).

- Determine Speed Changes

When acceleration and velocity have the same sign, the particle speeds up. When they have different signs, the particle slows down. It’s observed that particle slows down for \(0 < t < 2\) and speeds up for \(2 < t < 11/3\) and \(11/3 < t\).

- Determine Time to Reach Point (5,2)

Substitute \(x(t) = 5\) in the position function and solve for \(t\). The particle is at position \((5,2)\) at \(t≈1.35\) s.

- Describe Velocity and Speed

Velocity determines the rate of change of the position as a function of time and the direction of motion of the particle. It can be calculated as the derivative of the position function given by \(v(t) = 6t^{2} - 26t + 22\). Speed is the absolute of velocity and signifies how fast an object is moving without considering the direction.

Key concepts

Velocity and Acceleration

In calculus, when studying motion, velocity is the rate at which an object's position changes over time, and is given by the derivative of the position function with respect to time. In our exercise, the velocity function is determined by differentiating the position function, which results in v(t) = 6t^{2} - 26t + 22. This equation tells us how fast the particle is moving and in which direction at any point in time.

Acceleration, on the other hand, is the rate at which velocity changes over time. It is calculated by taking the derivative of the velocity function, yielding a(t) = 12t - 26. This tells us how the velocity of the particle is changing, which directly influences whether the particle is speeding up or slowing down. The signs of velocity and acceleration relative to each other provide insight into the particle's movement: when they share the same sign, the particle speeds up, and when they have opposite signs, the particle slows down.

Position Function

The position function is a foundational concept in calculus for describing motion. It represents the location of an object at a given time. In this problem, the position function is given by x(t) = 2t^{3} - 13t^{2} + 22t - 5, where t is the time in seconds. This cubic polynomial accounts for the distance traveled by the particle along the x-axis while staying on the line y=2. By plugging in different values of t into this function, we can determine the particle's exact position on the x-axis at various times. Notably, this function provides the mathematical foundation for finding velocity and acceleration by differentiation.

Direction of Motion

Understanding the direction of motion is crucial when analyzing particle motion. The direction is indicated by the sign of the velocity function: a positive value signifies that the particle is moving in the positive direction along the x-axis, to the right, and a negative value indicates that it's moving to the left. From the information obtained in this problem, we can deduce that when 0 <= t < 2 and 11/3 < t, the particle is moving to the right because the velocity is positive. Conversely, between t = 2 and t = 11/3, our particle reverses direction and moves to the left due to negative velocity. The changes in direction, crucially, occur at the points where the velocity is zero.

Calculus in Physics

Calculus is an essential tool in physics for modeling and solving problems related to motion and change. In physics, it allows us to describe how quantities such as position, velocity, and acceleration vary over time. For instance, in our particle motion example, we utilize differentiation, an operation in calculus, to find both velocity and acceleration from the position function. This example clearly demonstrates how calculus can translate a complex physical phenomenon into a set of manageable mathematical equations, making it possible to predict the future motion of an object or analyze its past behavior. Calculus bridges the gap between abstract mathematics and tangible physical concepts.