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Resistive forces are forces that oppose the motion of an object through a medium, such as air or water. These forces are essential in analyzing real-world scenarios where ideal conditions (like frictionless surfaces) do not exist. In the realm of Physics C: Mechanics, resistive forces help explain the behavior of objects in motion, particularly when external forces are balanced by opposing forces.
Air drag, a type of resistive force, occurs when an object moves through air, experiencing a force opposite to its direction of motion. This force depends on several factors:
The equation representing air drag is: $$ F_d = \frac{1}{2} \rho v^2 C_d A $$
Terminal velocity is the constant speed attained by an object when the net force acting upon it becomes zero. At this point, the downward gravitational force equals the upward resistive force, resulting in no further acceleration.
Mathematically, terminal velocity ($v_t$) can be derived by setting the gravitational force equal to the air drag: $$ mg = \frac{1}{2} \rho v_t^2 C_d A $$ Solving for $v_t$, we get: $$ v_t = \sqrt{\frac{2mg}{\rho C_d A}} $$
Where:
Several factors influence terminal velocity, including:
Understanding terminal velocity has practical applications in various fields:
Newton's Second Law states that the net force ($F_{net}$) acting on an object is equal to the mass ($m$) of the object multiplied by its acceleration ($a$): $$ F_{net} = ma $$ At terminal velocity, acceleration is zero ($a = 0$), so: $$ F_{net} = 0 $$ This implies that the gravitational force is balanced by the air drag: $$ mg = \frac{1}{2} \rho v_t^2 C_d A $$ Solving for terminal velocity ($v_t$): $$ v_t = \sqrt{\frac{2mg}{\rho C_d A}} $$
A typical velocity vs. time graph for an object falling under gravity with air resistance shows that velocity increases initially but approaches a constant value—terminal velocity—as time progresses. The graph illustrates how the forces balance out over time, leading to steady motion.

At terminal velocity, the kinetic energy input from gravity is balanced by the energy dissipated due to air drag. This balance ensures that the object's speed remains constant, as there's no net work being done to accelerate it further.
Consider a skydiver with a mass of $80\,kg$, a drag coefficient ($C_d$) of $1.0$, and a cross-sectional area ($A$) of $0.7\,m^2$. Assuming the air density ($\rho$) is $1.225\,kg/m^3$ and acceleration due to gravity ($g$) is $9.81\,m/s^2$, the terminal velocity ($v_t$) can be calculated as follows:
$$ v_t = \sqrt{\frac{2 \times 80\,kg \times 9.81\,m/s^2}{1.225\,kg/m^3 \times 1.0 \times 0.7\,m^2}} $$
$$ v_t = \sqrt{\frac{1569.6\,kg \cdot m/s^2}{0.8575\,kg/m \cdot s^2}} $$
$$ v_t = \sqrt{1826.05\,m^2/s^2} $$
$$ v_t \approx 42.73\,m/s $$
While the terminal velocity model provides valuable insights, it has certain limitations:
The Reynolds number ($Re$) is a dimensionless quantity used to predict flow patterns in different fluid flow situations. It helps determine whether the flow around an object is laminar or turbulent, which in turn affects the drag coefficient ($C_d$). $$ Re = \frac{\rho v D}{\mu} $$ Where:
Different flow regimes influence terminal velocity by altering the drag coefficient. For instance, at low Reynolds numbers, flow is laminar, and $C_d$ tends to be higher, whereas at high Reynolds numbers, flow becomes turbulent, often resulting in lower $C_d$ values.
As altitude increases, air density ($\rho$) decreases, leading to reduced air drag. Consequently, an object's terminal velocity increases with altitude. This phenomenon is evident in high-altitude skydives, where jumpers can achieve higher terminal velocities compared to lower altitudes.
Temperature influences air density and viscosity, thereby affecting air drag. Warmer air is less dense, resulting in lower air drag and higher terminal velocity. Conversely, colder air increases air drag, reducing terminal velocity.
Aspect | Air Drag | Terminal Velocity |
Definition | Resistive force opposing an object's motion through air. | Constant velocity achieved when gravitational force equals air drag. |
Dependence on Velocity | Proportional to $v^2$. | Determined by the balance of forces, independent of further acceleration. |
Key Equation | $F_d = \frac{1}{2} \rho v^2 C_d A$ | $v_t = \sqrt{\frac{2mg}{\rho C_d A}}$ |
Factors Influencing | Velocity, cross-sectional area, air density, drag coefficient. | Mass, gravitational acceleration, air density, drag coefficient, cross-sectional area. |
Real-World Examples | Falling objects, projectiles, vehicles in motion. | Skydivers reaching terminal speed, parachute descent. |
To master terminal velocity, remember the mnemonic V = √(2mg / (ρCdA)) where each variable plays a key role. Practice deriving terminal velocity from Newton's Second Law to reinforce your understanding. When tackling AP exam problems, always sketch free-body diagrams to visualize forces. Additionally, familiarize yourself with different drag coefficients for various shapes to quickly identify $C_d$ values during calculations.
Did you know that the concept of terminal velocity was first explored by Galileo Galilei? Additionally, some birds, like the peregrine falcon, can adjust their terminal velocity mid-flight by changing their body orientation, allowing them to dive at incredible speeds. Understanding terminal velocity isn't just academic—it plays a crucial role in designing safe spacecraft re-entry profiles and enhancing the performance of high-speed trains by minimizing air resistance.
Many students confuse air drag with simple friction, neglecting its dependence on velocity squared. For instance, incorrectly applying linear friction formulas to air drag problems can lead to inaccurate results. Another common mistake is overlooking the role of the drag coefficient ($C_d$) when calculating terminal velocity, which can significantly affect the outcome. Lastly, assuming constant air density at all altitudes may simplify calculations but introduces errors, especially in high-altitude scenarios.