Question

Air resistance (adapted from Putnam Exam, 1939 ) An object moves freely in a straight line, acted on by air resistance, which is proportional to its speed; this means its acceleration is \(a(t)=-k v(t) .\) The velocity of the object decreases from \(1000 \mathrm{ft} / \mathrm{s}\) to \(900 \mathrm{ft} / \mathrm{s}\) over a distance of \(1200 \mathrm{ft}\). Approximate the time required for this deceleration to occur. (Exercise 50 may be useful.)

Short answer

To summarize, in order to approximate the time required for the object to decelerate from 1000 ft/s to 900 ft/s while covering a distance of 1200 ft, we can use the expression: $$ \Delta t \approx \frac{1}{k}\ln(1.1) $$ where \(\Delta t \approx 0.1/k\), and \(k\) is a constant related to the problem that cannot be determined with the given information. The time required for this deceleration to occur is approximately one-tenth of the reciprocal of the constant \(k\).

Step-by-step solution

Write down the governing equation

The problem states that the acceleration is proportional to the velocity: $$ a(t) = -k v(t) $$

Solve the differential equation

We can rewrite the equation as: $$ \frac{dv}{dt} = -k v(t) $$ Now, separate the variables: $$ \frac{dv}{v(t)} = -k dt $$ Integrate both sides: $$ \int \frac{dv}{v(t)} = -k \int dt $$ Which leads to: $$ \ln |v(t)| = -kt + C_1 $$

Solve for v(t)

Exponentiate both sides to solve for the velocity as a function of time. $$ v(t) = e^{-kt + C_1} = e^{-kt} e^{C_1} $$ Let \(e^{C_1} = A\), then we get: $$ v(t) = Ae^{-kt} $$ Here, \(A\) and \(k\) are constants to be determined with given conditions.

Apply the initial condition

At \(t=0\), the initial velocity is \(v(0) = 1000 \text{ ft/s}\), so we can calculate \(A\) by: $$ 1000 = Ae^{-k(0)} $$ Thus, $$ A = 1000 $$ So the velocity function becomes: $$ v(t) = 1000e^{-kt} $$

Write down the equation for distance

We can now find the position function \(s(t)\) by integrating the velocity function: $$ s(t) = \int v(t) dt = \int 1000e^{-kt} dt $$ After integrating, we have: $$ s(t) = -\frac{1000}{k}e^{-kt} + C_2 $$

Apply the distance condition

The object covers a distance of 1200 ft as its speed reduces from 1000 ft/s to 900 ft/s. So, when \(\Delta s = 1200\text{ ft}\), \(\Delta v = -100 \text{ ft/s}\). We can express this as: $$ s(t_2) - s(t_1) = -\frac{1000}{k}e^{-kt_2} + C_2 - (-\frac{1000}{k}e^{-kt_1} + C_2) = 1200 $$ Which simplifies to: $$ \frac{1000}{k}(e^{-kt_1} - e^{-kt_2}) = 1200 $$ Now, substitute the velocities \(v(t_1)=1000 \text{ ft/s}\) and \(v(t_2)=900 \text{ ft/s}\), and plug in the velocity expression: $$ \frac{1000}{k}(e^{-kt_1} - e^{-kt_2}) = 1000 - 900 $$

Solve for time

The equation becomes: $$ e^{-kt_1} - e^{-kt_2} = 1 - \frac{900}{1000} $$ Then use: $$ e^{-kt} = \frac{v(t)}{1000} $$ So we get: $$ t_2 = t_1 + \frac{1}{k}\ln\left(\frac{1000}{100 \cdot 9}\right) \approx t_1 + \frac{1}{k}\ln(1.1) $$ Now we can approximate the time required for this deceleration to occur: $$ \Delta t = t_2 - t_1 \approx \frac{1}{k}\ln(1.1) $$ Unfortunately, we cannot find the exact value of \(k\) with the given information. However, we can observe that \(\ln(1.1) \approx 0.1\), then \(\Delta t \approx \frac{1}{k} (0.1)\). From this approximation, we know that the time required for the deceleration to occur is approximately one-tenth of the reciprocal of the constant \(k\).

Key concepts

Differential Equations

In the world of physics and mathematics, differential equations are tools that help describe relationships involving rates of change. In this problem, the differential equation given is: \[ a(t) = -k v(t) \] It shows that the acceleration of the object is negatively proportional to its velocity due to air resistance. A differential equation is simply an equation that involves a function and its derivatives.
In this scenario, we have a first-order differential equation because it relates the first derivative of velocity, \( \frac{dv}{dt} \), to velocity itself. To solve it, we separated the variables and integrated both sides to find how velocity changes over time. This gives us insight into how the speed of the object decreases as it moves through the air.
Differential equations like this one are instrumental in modeling real-world phenomena, from weather patterns to electrical circuits.

Velocity

Velocity describes how fast an object moves and in which direction. In this problem, initial velocity and its reduction due to air resistance need to be understood. The object starts with a velocity of 1000 ft/s and slows down to 900 ft/s.
The relationship between velocity and acceleration due to air resistance is encapsulated in our differential equation. Solving the equation through integration gives us the velocity function:\( v(t) = 1000e^{-kt} \) Here, \( v(t) \) is the velocity at any time \( t \). The function shows an exponential decay, meaning the object slows down exponentially over time because of the friction with the air.
Understanding this decreasing velocity is essential for predicting how objects behave when impacted by forces like air resistance.

Acceleration

Acceleration is how quickly the velocity of an object changes, affected by forces like air resistance in our scenario. The given equation is\( a(t) = -k v(t) \), showcasing that acceleration is directly impacted by velocity and a negative constant \( k \).
Since acceleration is proportional to velocity, the faster the object moves, the greater the opposing force acting on it. This is indicative of a deceleration due to air resistance which naturally happens to moving objects.
Incorporating air resistance, the acceleration is negative, showing the retarding force. As velocity decreases, the magnitude of acceleration decreases too because less force is needed to maintain this deceleration over time. This relationship explains why lighter objects fall slower compared to heavier ones when air resistance is considered.
It's vital to consider acceleration in motion problems since it directly affects how and when the object will stop, slow down, or speed up.

Integration

Integration is a key mathematical operation used to find accumulative quantities, like the distance traveled from velocity or velocity from acceleration. In this exercise, we integrated the separated form of our differential equation to find the resulting function for velocity: \[ \int \frac{dv}{v(t)} = -k \int dt \] Solving this integral gave us \( \ln |v(t)| = -kt + C_1 \), which was then further simplified to our velocity function.
Integration was also used to determine the position of the object over time, moving from velocity to distance. When we performed:\( s(t) = \int v(t) dt \), it results in the position function that enabled us to calculate the distance problem described.
Thus, integration acts as a bridge, allowing us to move from rates of change back to actual amounts—quantities we can measure, like distance and velocity.

Exponential Function

Exponential functions describe phenomena where growth or decay happens at a continuous rate. In our exercise, solving the differential equation for velocity leads to:\( v(t) = Ae^{-kt} \).
This is an exponential decay function, indicating that velocity decreases rapidly at first but more slowly over time. The constant \( -k \) determines the rate of this decay; a larger \( k \) means quicker reduction.
Exponential functions are common in natural processes like radioactive decay or population decline, where changes decrease over time. Understanding how to handle them through calculations like integration is crucial in applying mathematics to everyday life scenarios.
In essence, this exercise uses an exponential function to model how velocity decreases as the object is subjected to air resistance, a process observable in various real-world settings.