Question

The falling object in Example 2 satisfies the initial value problem $$ d v / d t=9.8-(v / 5), \quad v(0)=0 $$ (a) Find the time that must elapse for the object to reach \(98 \%\) of its limiting velocity. (b) How far does the object fail in the found in part (a)?

Short answer

Answer: The object falls approximately 437.17 meters.

Step-by-step solution

Solve the initial value problem

To find the formula for the velocity as a function of time, we will need to solve the given differential equation: $$ \frac{d v}{d t}=9.8-\frac{v}{5} $$ This is a first-order linear differential equation, and we can solve it using an integrating factor. First, rewrite the equation in the standard form: $$ \frac{d v}{d t}+\frac{1}{5}v=9.8 $$ The integrating factor is given by: $$ \mu(t)=e^{\int\frac{1}{5}dt}=e^{\frac{1}{5}t} $$ Multiplying both sides of the equation by the integrating factor, we obtain: $$ e^{\frac{1}{5}t}\frac{d v}{d t}+\frac{1}{5}e^{\frac{1}{5}t}v=9.8e^{\frac{1}{5}t} $$ This is now an exact differential equation, in the form: $$ \frac{d}{d t}\left(v(t)e^{\frac{1}{5}t}\right)=9.8e^{\frac{1}{5}t} $$ Integrate both sides with respect to t: $$ v(t)e^{\frac{1}{5}t}=\int9.8e^{\frac{1}{5}t}dt $$ We can solve the integration on the right side using integration by substitution: Let \(u=\frac{1}{5}t\), so \(du=\frac{1}{5}dt\) and \(dt=5du\). Then, $$ v(t)e^{\frac{1}{5}t}=49\int e^{u}du=49e^{u}+C=49e^{\frac{1}{5}t}+C $$ Now, multiply both sides by \(e^{-\frac{1}{5}t}\) to isolate \(v(t)\): $$ v(t)=49e^{-\frac{1}{5}t}+Ce^{-\frac{1}{5}t} $$ To find the constant C, use the initial condition \(v(0)=0\): $$ 0=49+5C\Rightarrow C=-9.8 $$ So, our formula for the velocity as a function of time is: $$ v(t)=49\left(e^{-\frac{1}{5}t}-e^{-1}\right) $$

Find the time needed to reach 98% of the limiting velocity

The limiting velocity occurs as \(t\rightarrow\infty\). In this limit, \(e^{-\frac{1}{5}t} \rightarrow 0\), so the limiting velocity is: $$ \lim_{t \to \infty}v(t)=49\left(0-e^{-1}\right)=-49e^{-1} $$ Now, we need to find the time when the velocity is 98% of the limiting velocity: $$ 0.98(-49e^{-1})=49\left(e^{-\frac{1}{5}t}-e^{-1}\right) $$ To find \(t\), we can solve the above equation: $$ 0.98e^{-1}+e^{-1}=e^{-\frac{1}{5}t}\\ 1.98e^{-1}=e^{-\frac{1}{5}t} $$ Take the natural logarithm of both sides: $$ -\frac{1}{5}t = \ln\left(1.98e^{-1}\right) $$ Now, solve for \(t\): $$ t=-5\ln\left(1.98e^{-1}\right)\approx 29.27 \text{ seconds} $$ So, it takes approximately 29.27 seconds for the object to reach 98% of its limiting velocity.

Find the distance fallen during this time

To find the distance fallen during the first 29.27 seconds, we need to integrate the velocity formula with respect to time: $$ s(t)=\int v(t) dt = \int 49\left(e^{-\frac{1}{5}t}-e^{-1}\right) dt $$ Integrating, we get: $$ s(t)=-245e^{-\frac{1}{5}t}+49te^{-1}+C $$ Since we're considering the falling object, we know that the initial distance fallen is 0, which means \(s(0)=0\). Using this, we can find C: $$ 0=-245e^{0}+49(0)e^{-1}+C \Rightarrow C=245 $$ So, the distance fallen as a function of time is: $$ s(t)=-245e^{-\frac{1}{5}t}+49te^{-1}+245 $$ Now, we can find the distance fallen during the first 29.27 seconds: $$ s(29.27)=-245e^{-\frac{1}{5}(29.27)}+49(29.27)e^{-1}+245\approx 437.17 \text{ meters} $$ So, the object falls approximately 437.17 meters during the time it takes to reach 98% of its limiting velocity.

Key concepts

Understanding Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are a powerful tool in expressing the relationship between varying quantities and their rates of change. For instance, when an object falls towards the ground, its velocity changes over time due to gravity and air resistance. This type of motion can be modeled using a differential equation, which in our exercise is \( \frac{dv}{dt} = 9.8 - \frac{v}{5} \).

The left side of this equation, \( \frac{dv}{dt} \), represents the rate of change of velocity (acceleration) with respect to time, while the right side provides the net force acting on the object divided by its mass, according to Newton's second law. The constant 9.8 is the acceleration due to gravity (approximately in meters per second squared on Earth's surface) and the term \( \frac{v}{5} \) represents the opposing force due to air resistance proportional to the velocity.

In solving differential equations, we often look for a function that describes the velocity over time, \( v(t) \), that satisfies the given initial condition. In this case, the initial condition is \( v(0)=0 \), meaning that the object starts from rest. This example shows how differential equations can predict complex real-world behaviors from just a few simple principles.

The Role of an Integrating Factor in Solving Differential Equations

An integrating factor is a functional tool designed to simplify the process of solving linear differential equations. To understand this concept, it's beneficial to recognize that not all differential equations can be solved by simple integration. The integrating factor technique is particularly useful for first-order linear ordinary differential equations, which are of the form \( \frac{dy}{dx} + P(x)y = Q(x) \).

In our exercise, we encounter such an equation: \( \frac{dv}{dt} + \frac{1}{5}v = 9.8 \). The goal is to rewrite it into an exact differential that can be integrated directly. To achieve this, we multiply the entire equation by the integrating factor \( \mu(t) = e^{\int\frac{1}{5}dt} = e^{\frac{1}{5}t} \), which is derived from the function \( P(x) \), in this case, \( \frac{1}{5} \). Multiplying through by the integrating factor, we transform the equation into a form that allows for straightforward integration.

After integration and applying the initial condition \( v(0)=0 \), we derive the function for velocity over time. The integrating factor aids in transforming a seemingly tricky differential equation into one that can be integrated, showcasing the elegance of the method. It's a vital process in the toolkit of anyone working with differential equations, illuminating how mathematics can tame the complexity of dynamic systems.

Approaching the Limiting Velocity

Limiting velocity, or terminal velocity, is a concept in physics describing the maximum speed an object can reach when falling through a fluid (e.g., air or water) when the net force acting on it is zero. This happens when the force due to air resistance equals the gravitational pull, resulting in zero acceleration. In the initial value problem provided, the velocity function \( v(t) \) eventually approaches a steady state as time tends to infinity.

To find the limiting velocity, one would evaluate the limit \( \lim_{t \to \infty}v(t) \), which represents the theoretical maximum speed the object would reach under the force of gravity and the resistance of the medium it's falling through. In our exercise, as \( t \) approaches infinity, the exponential term \( e^{-\frac{1}{5}t} \) in our velocity function goes to zero, and we are left with the limiting velocity value.

This gives us insight into the behavior of objects under constant forces and resistance, and allows us to calculate important predictive properties, like the time required for the object to reach a certain percentage of this maximum speed—98% in the problem's context. Understanding the concept of limiting velocity is essential not just in theoretical physics, but also in practical applications such as designing parachutes or estimating the speed at which raindrops fall.