91Ó°ÊÓ

The position function tells us the location of a body at any given time. In this exercise, the position function is given as a function of time, expressed as \( x(t) = \frac{t^3}{3} \). This means as time progresses, the position of the body changes according to this cubic relationship.
This function is crucial since it allows us to track the movement of the body over time.
Every value of \( t \) plugged into this function gives the corresponding value of \( x \), which is the body's position in meters.
For example, when \( t = 2 \text{ sec} \), \( x(2) = \frac{2^3}{3} = \frac{8}{3} \text{ meters} \). This lets us understand not only where the body is at different times, but also how it moves.

Differentiation is a fundamental concept in calculus used to find the rate of change of functions. Here, we use differentiation to determine the velocity from the position function. The position function \( x(t) = \frac{t^3}{3} \) is differentiated with respect to time \( t \) to get the velocity function.
Applying the differentiation rule \( \frac{d}{dt}(t^n) = nt^{n-1} \), we get:

• \( \frac{d}{dt}\left( \frac{t^3}{3} \right) = t^2 \)

Thus, the velocity function is \( v(t) = t^2 \), indicating how position changes over time.
The ability to differentiate allows us to move from knowing where an object is over time to understanding how its speed changes.

The calculation of velocity is essential to understanding motion. Here, velocity is calculated from the differentiated position function, giving us \( v(t) = t^2 \). To find specific velocities at given times, you simply substitute the time values into this function.

• At \( t = 0 \text{ sec} \), the velocity is \( v(0) = 0^2 = 0 \, \text{m/s} \).
• At \( t = 2 \text{ sec} \), the velocity becomes \( v(2) = 2^2 = 4 \, \text{m/s} \).

These calculations provide the initial and final velocities needed to evaluate the changes in the kinetic energy.
This step bridges the gap from static position information to dynamic motion analysis.

Kinetic energy is the energy a body possesses due to its motion. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.
It is calculated with the formula:

• \( W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \)

Here, \( m \) is the mass, \( v_i \) is the initial velocity, and \( v_f \) is the final velocity. With a mass of \( 2 \, \text{kg} \), an initial velocity of \( 0 \, \text{m/s} \) and a final velocity of \( 4 \, \text{m/s} \), you get:

• \( W = \frac{1}{2} \times 2 \times (4^2) - \frac{1}{2} \times 2 \times (0^2) = 16 \, \text{J} \)

This change in kinetic energy shows the amount of work done by the force acting on the body.
Understanding this principle is critical for solving many physics problems where forces and motion are involved.