91Ó°ÊÓ

The position function, denoted as \(x(t)\), describes how the position of an object changes over time. In this exercise, we are given the position function of a particle-like object as a polynomial in terms of time \(t\):

\[ x(t) = (3 \text{ m/s}) t - (4 \text{ m/s}^2) t^2 + (1 \text{ m/s}^3) t^3 \]

This equation tells us that the position depends on three components: a linear term, a quadratic term, and a cubic term. Each term represents a contribution to the position based on different powers of time.

Understanding the position function is crucial because it allows us to find other physical quantities, such as velocity and acceleration, by differentiating the function with respect to time.

To find the velocity function \(v(t)\), we differentiate the position function \(x(t)\) with respect to time. The differentiation process gives us the rate at which the position is changing over time, which is the velocity:

\[ v(t) = \frac{d}{dt}(x(t)) = \frac{d}{dt} [(3 \text{ m/s}) t - (4 \text{ m/s}^2) t^2 + (1 \text{ m/s}^3) t^3] \]

When we perform the differentiation, we get:

\[ v(t) = 3 \text{ m/s} - 8 \text{ m/s}^2 t + 3 \text{ m/s}^3 t^2 \]

This velocity function gives us the speed at any point in time \(t\). Knowing the velocity at specific times is essential for numerous calculations, including determining the work done by a force over a specified time interval.

Kinetic energy \(K\) is the energy that an object possesses due to its motion. The formula for kinetic energy is:

\[ K = \frac{1}{2} m v^2 \]

Where \(m\) is the mass of the object and \(v\) is its velocity. In this exercise, we use the Work-Energy Theorem, which states that the work \(W\) done on an object is equal to the change in kinetic energy \(\Delta K\):

\[ W = \frac{1}{2} m \big[ v(t_2)^2 - v(t_1)^2 \big] \]

Let's compute the kinetic energies at the given times. At \(t_1 = 0.0 \text{ s}\), the velocity is \(v(0) = 3 \text{ m/s}\):

\[ K_1 = \frac{1}{2} \times 3.0 \text{ kg} \times (3 \text{ m/s})^2 = 13.5 \text{ J} \]

At \(t_2 = 4.0 \text{ s}\), the velocity is \(v(4) = 19 \text{ m/s}\):

\[ K_2 = \frac{1}{2} \times 3.0 \text{ kg} \times (19 \text{ m/s})^2 = 541.5 \text{ J} \]

The work done from \(t_1\) to \(t_2\) is the change in kinetic energy:

\[ W = K_2 - K_1 = 541.5 \text{ J} - 13.5 \text{ J} = 528 \text{ J} \]

Thus, the work done on the object by the force from \(t_1 = 0.0 \text{ s}\) to \(t_2 = 4.0 \text{ s}\) is \(528 \text{ J}\).