91Ó°ÊÓ

Differential equations play a crucial role in modeling dynamic systems in physics. A differential equation involves functions and their derivatives, illustrating how the rate of change of a quantity is linked to the quantity itself. In our example, we deal with the change in an object's velocity over time.
The problem involves a resistive force and gravity affecting velocity. The resistive force depends on velocity, represented as \( F = -bv \), where \( b \) is a constant. Gravity exerts a constant force \( mg \), making the net force on the object: \( ma = mg - bv \), where \( a \) is the acceleration.
By expressing acceleration as \( \frac{dv}{dt} \), the net force becomes a first-order differential equation:

• \( m\frac{dv}{dt} = mg - bv \)

This equation needs to be rearranged to separate variables \( v \) and \( t \), allowing each side to be integrated independently.

Newton's second law of motion is fundamental to understanding the behavior of objects under force. It states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration: \( F = ma \). This is the core equation for analyzing motion under various forces.
In our exercise, the object is influenced by two main forces: gravity and a resistive force proportional to velocity. By applying Newton's second law, we combine these forces:

• The gravitational force \( F_g = mg \)
• The resistive force \( F_r = -bv \)

The total force acting on the object becomes \( F = mg - bv \). Using \( ma = mg - bv \) sets up the differential equation required to model the object's motion. This relationship connects the forces to the object's mass and velocity, allowing us to predict velocity over time by solving the equation.

To predict an object's velocity over time when subject to forces, you derive a formula from the original differential equation. The process involves separating variables and integrating both sides. Initially, the differential equation \( m\frac{dv}{dt} = mg - bv \) is manipulated to prepare for integration by dividing by \( mg - bv \):

• \( \frac{dv}{mg - bv} = \frac{dt}{m} \)

Upon integration, the left-hand side requires substitution, resulting in a logarithmic function.

• \( \int \frac{dv}{mg - bv} = -\frac{1}{b} \ln |mg - bv| + C \)

The right-hand side integrates into a simple linear function \( \frac{t}{m} \). The full integration yields an equation involving \( e^{-bt/m} \) terms, reflecting how velocity approaches a terminal value \( \frac{mg}{b} \), factoring in initial velocity. Adjusting for initial conditions, two cases arise:

• With downward initial velocity: \( v(t) = \frac{mg}{b} \left(1 - e^{-bt/m}\right) + v_0 e^{-bt/m} \)
• With upward initial velocity: \( v(t) = \frac{mg}{b} \left(1 - e^{-bt/m}\right) - v_0 e^{-bt/m} \)

These formulas illustrate how velocity changes with time under both resistive and gravitational forces.