Question

One possible model that describes the free fall of an object in a gravitational field subject to air resistance uses the equation \(v^{\prime}(t)=g-b v,\) where \(v(t)\) is the velocity of the object for \(t \geq 0\), \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(b>0\) is a constant that involves the mass of the object and the air resistance. a. Verify by substitution that a solution of the equation, subject to the initial condition \(v(0)=0,\) is \(v(t)=\frac{g}{b}\left(1-e^{-b t}\right)\). b. Graph the solution with \(b=0.1 s^{-1}\). c. Using the graph in part (c), estimate the terminal velocity \(\lim _{t \rightarrow \infty} v(t)\).

Short answer

Question: Verify the given solution \(v(t) = \frac{g}{b}(1 - e^{-bt})\) for the differential equation \(v'(t) = g - bv.\) Then, graph the solution using \(b=0.1s^{-1}\) and estimate the terminal velocity. Answer: We have verified the given function as a solution to the differential equation by substitution. The graph of the solution with \(b=0.1s^{-1}\) has the equation \(v(t) = 98(1 - e^{-0.1t}).\) Upon analyzing the graph, we can estimate the terminal velocity to be approximately 98 m/s.

Step-by-step solution

(a) Verify the Solution by Substitution

To verify that the function \(v(t) = \frac{g}{b}(1 - e^{-bt})\) is a solution to the differential equation \(v'(t) = g - bv,\) we will compute the derivative of \(v(t)\) and substitute it into the given equation. 1. Compute the derivative of \(v(t)\): \(v'(t) = \frac{d}{dt} \left(\frac{g}{b}(1 - e^{-bt})\right) = (-g e^{-bt})\) 2. Substitute the derivative into the given equation: \(g - bv = g - b\left(\frac{g}{b}(1 - e^{-bt})\right) = g - g(1 - e^{-bt})\) 3. Check if both sides are equal: \((-g e^{-bt}) = g - g(1 - e^{-bt})\). The equation holds true, so the given function is a solution to the differential equation with the initial condition \(v(0) = 0.\)

(b) Graph the Solution with b=0.1s^{-1}

To graph the given equation with \(b = 0.1s^{-1},\) we'll rewrite the function using this value and plot it: \(v(t) = \frac{9.8}{0.1}(1 - e^{-0.1t}) = 98(1 - e^{-0.1t})\) Now plot the function to obtain the graph of the solution.

(c) Estimate the Terminal Velocity

We are asked to estimate the terminal velocity, which is given by the limit of the function \(v(t)\) as \(t \rightarrow \infty\): \(\lim _{t \rightarrow \infty} v(t) = \lim_{t \rightarrow \infty} 98(1 - e^{-0.1t})\) To estimate the limit, consider that the exponential term \(e^{-0.1t}\) approaches 0 as \(t\) approaches infinity: \(\lim_{t \rightarrow \infty} e^{-0.1t} = 0\) Using this result, we obtain the terminal velocity: \(\lim _{t \rightarrow \infty} v(t) = 98(1 - 0) = 98 \, \mathrm{m}/\mathrm{s}\) So the terminal velocity is approximately 98 m/s.

Key concepts

Air Resistance

When an object moves through the air, it experiences a force known as air resistance. Air resistance acts against the motion of the object, slowing it down. This force is sometimes called drag and is influenced by several factors:

• The object's speed.
• Its cross-sectional area.
• The air density.
• The shape of the object.

In the context of free fall motion, air resistance becomes increasingly important as the speed of the object increases. In many models, air resistance is proportional to the velocity of the object, leading to a different outcome than when only gravity is considered.
This is represented by the equation: \[ v'(t) = g - bv \] where \(b\) is a positive constant representing the strength of the air resistance. The higher the value of \(b\), the stronger the air resistance.

Free Fall Motion

Free fall motion describes the movement of an object solely under the influence of gravity. In the ideal case of vacuum conditions, where there is no air resistance, objects of different masses fall at the same rate due to gravity. This is approximately \(9.8 \, \text{m/s}^2\) near the Earth's surface.However, in reality, air plays a significant role. When air resistance is considered, the motion becomes more complex than simple acceleration under gravity. The governing equation for free fall with air resistance assumes that the resistive force is proportional to the velocity.
The differential equation \[ v'(t) = g - bv \] describes how the velocity of the object changes over time. At the beginning of the motion, air resistance is minimal, and the velocity increases rapidly. As time progresses, air resistance grows, and the acceleration decreases, leading to the object reaching a steady speed, known as terminal velocity.

Terminal Velocity

Terminal velocity is the constant speed that a freely falling object eventually reaches when the force of air resistance equals the gravitational force acting on it. At this speed, the acceleration of the object is zero because the net force is zero.In the context of our differential equation, terminal velocity can be found by evaluating the limit of \(v(t)\) as time approaches infinity. The exponential term in the equation, \[ v(t) = \frac{g}{b}(1 - e^{-bt}) \] becomes negligible over time, so
\( e^{-bt} \rightarrow 0 \). Therefore, the terminal velocity, \( v_t \), is given by \[ v_t = \frac{g}{b} \].This means that the terminal velocity depends on both gravitational acceleration and the air resistance coefficient \(b\). The specific case of \(b = 0.1 \, \text{s}^{-1}\) leads to a terminal velocity of \(98 \, \text{m/s}\), as calculated from the given equation. This steady speed signifies the balance between gravity pulling the object downwards and air resistance pushing upwards.