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In physics, a position function describes how the location of an object changes over time. Here, the position function is given as \( x(t) = 2t^4 + 5t + 4 \). This function tells us the position \( x \) of a body at any particular time \( t \). Each term in the function captures different aspects of motion. The term \( 2t^4 \) depicts a rapidly increasing component as time progresses, pointing to acceleration. The term \( 5t \) suggests linear movement, contributing to constant velocity, while the constant \( 4 \) indicates the initial position where the body started off. Understanding this function provides the foundation needed to study velocity and kinetic energy as time evolves.

Velocity is the rate of change of position with time, and it highlights how fast an object is moving and in what direction. To craft a velocity function from a position function, we simply differentiate. Differentiation, in this context, means finding out how one quantity changes in relation to another. In our case, it involves differentiating the position function with respect to time \( t \).The original position function is \( x(t) = 2t^4 + 5t + 4 \).After differentiation, it becomes \( v(t) = \frac{d}{dt}(2t^4 + 5t + 4) = 8t^3 + 5 \). This newly derived equation \( v(t) \) now gives us the velocity at any point in time. By plugging in a value for \( t \), one can readily find out how fast the object is moving at any specific instant.

Differentiation is a fundamental technique in calculus used to determine the rate at which a quantity changes. In simple terms, it helps you find the slope or steepness of a function at any given point. Differentiating the position function \( x(t) = 2t^4 + 5t + 4 \) with respect to time \( t \), we get its derivative, the velocity function: \( v(t) = 8t^3 + 5 \).Here's a step-by-step approach:

• The derivative of \( t^n \) is \( nt^{n-1} \). Hence, the derivative of \( 2t^4 \) becomes \( 8t^3 \), because you multiply the exponent by the coefficient and then subtract one from the exponent.
• The derivative of \( 5t \) is simply \( 5 \), as the power of \( t \) is 1, which gives \( 5 \times 1 \times t^0 \) (\( t^0 = 1 \)).
• Constants, such as \( 4 \), disappear when differentiated because they don’t change with time.

These elements together form the velocity function. Thus, differentiation is not only about finding derivatives but understanding the change of quantities.

Kinetic energy is the energy an object possesses due to its motion. The formula for kinetic energy \( KE \) is \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. This calculation includes important steps:1. **Initial Kinetic Energy:** - At \( t = 0 \), the velocity \( v(0) = 5 \) m/s. Thus, the initial kinetic energy is calculated as: \[ KE_{\text{initial}} = \frac{1}{2} \times 2 \times (5)^2 = 25 \, \text{J} \]2. **Kinetic Energy at 1 second:** - At \( t = 1 \), the velocity \( v(1) = 13 \) m/s. The kinetic energy then is: \[ KE_{\text{final}} = \frac{1}{2} \times 2 \times (13)^2 = 169 \, \text{J} \]3. **Increase in Kinetic Energy:** - The increase in kinetic energy is simply the difference between kinetic energies at \( t = 1 \) second and \( t = 0 \) seconds: \[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 169 \, \text{J} - 25 \, \text{J} = 144 \, \text{J} \]This calculation illustrates how the change in velocity over time leads to a considerable increase in the body's kinetic energy.