Question

Acceleration, velocity, position Suppose the acceleration of an object moving along a line is given by \(a(t)=-k v(t),\) where \(k\) is a positive constant and \(v\) is the object's velocity. Assume the initial velocity and position are given by \(v(0)=10\) and \(s(0)=0\) respectively. a. Use \(a(t)=v^{\prime}(t)\) to find the velocity of the object as a function of time. L. Use \(v(t)=s^{\prime}(t)\) to find the position of the object as a function of time. c. Use the fact that \(\frac{d v}{d t}=\frac{d v}{d s} \frac{d s}{d t}\) (by the Chain Rule) to find the velocity as a function of position.

Short answer

Based on the given information, we derived the following functions: a. The velocity as a function of time is given by the function: $$ v(t) = 10 e^{-kt}. $$ b. The position as a function of time is given by the function: $$ s(t) = -\frac{10}{k} e^{-kt} + \frac{10}{k}. $$ c. The velocity as a function of position is given by the function: $$ \frac{dv}{ds} = -10 k s'(t). $$

Step-by-step solution

a. Finding the velocity as a function of time

First, we need to find the velocity function \(v(t)\). We know the acceleration function \(a(t)\), and we can write the relationship between acceleration and velocity as \(a(t) = v'(t)\). So, we have: $$ v'(t) = -k v(t). $$ This is a first-order, separable differential equation. To solve it, we need to integrate both sides: $$ \int \frac{dv}{v(t)} = -k \int dt. $$ Now, we can do the integrations: $$ \ln(v(t)) = -kt + C_1. $$ To eliminate the logarithm, we can use the exponential function: $$ v(t) = e^{-kt + C_1}. $$ Now, let's use the initial condition \(v(0) = 10\): $$ 10 = e^{C_1} \Rightarrow C_1 = \ln(10). $$ So, the velocity function is: $$ v(t) = 10 e^{-kt}. $$

b. Finding the position as a function of time

Now, we need to find the position function \(s(t)\). We know the velocity function \(v(t)\), and we can write the relationship between position and velocity as \(s'(t) = v(t)\). So, we have: $$ s^\prime(t) = 10 e^{-k t}. $$ To obtain \(s(t)\), we integrate both sides: $$ \int ds = 10 \int e^{-kt} dt. $$ Now, we can do the integrations: $$ s(t) = -\frac{10}{k} e^{-kt} + C_2. $$ Using the initial condition \(s(0) = 0\): $$ 0 = -\frac{10}{k} + C_2 \Rightarrow C_2 = \frac{10}{k}. $$ So, the position function is: $$ s(t) = -\frac{10}{k} e^{-kt} + \frac{10}{k}. $$

c. Finding the velocity as a function of position

We are given that \(\frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt}\). Using the Chain Rule, we plug in the known functions of \(v(t)\) and \(s(t)\): $$ \frac{d}{dt}(10e^{-kt}) = \frac{dv}{ds}\frac{d}{dt}\left( -\frac{10}{k} e^{-kt} + \frac{10}{k} \right). $$ Since \(v(t) = \frac{ds}{dt}\), the equation becomes: $$ \frac{d}{dt}(10e^{-kt}) = \frac{dv}{ds} \cdot 10e^{-kt}. $$ Now, we divide both sides by \(10e^{-kt}\): $$ \frac{d}{d t}(e^{-kt}) = \frac{1}{10} \cdot \frac{dv}{ds}. $$ Also, we know that \(\frac{d}{d t}(e^{-kt}) = - k e^{-kt}\), so we have: $$ - k e^{-kt} = \frac{1}{10} \cdot \frac{dv}{ds}. $$ Now, we can get rid of the exponential term by substitution: $$ - k v = \frac{1}{10} \cdot \frac{dv}{ds}. $$ Now, we can solve for \(\frac{dv}{ds}\): $$ \frac{dv}{ds} = -10 k v. $$ So the velocity as a function of position is: $$ \frac{dv}{ds} = -10 k s'(t). $$

Key concepts

Understanding Acceleration

Acceleration is the rate at which an object's velocity changes over time. In this specific problem, the acceleration is defined by the equation \(a(t) = -k v(t)\). Here, \(k\) is a constant that influences the rate of deceleration since it is multiplied by the negative sign. The negative sign indicates that the acceleration is actually working in the direction opposite to the velocity, causing the object to slow down.

When solving problems involving acceleration, it's essential to recognize how it affects an object's motion. By understanding that acceleration can either increase or decrease depending on its relation to velocity, we can better predict the changes in movement over time. One key technique is to solve the differential equation \(v'(t) = -k v(t)\), which relates the velocity function to time.

• Acceleration is the change in velocity over time.
• It is influenced by constants such as \(k\).
• Negative acceleration, or deceleration, slows an object down.

Discovering Velocity

Velocity is a measure of how fast an object is moving in a specific direction. It can be affected by acceleration and, in this exercise, we see that the velocity function, \(v(t) = 10e^{-kt}\), is determined by solving a differential equation. Here, the exponential decay signifies the gradual reduction in velocity over time due to the negative acceleration.

To find the velocity as a function of time, we used the initial condition \(v(0) = 10\), which provides the starting speed of the object. This step is crucial because initial conditions help identify specific solutions in differential equations. Understanding how to manipulate and solve these equations allows us to predict the velocity at any given time.

• Velocity refers to the speed and direction of motion.
• Differential equations can define how velocity changes over time.
• Initial conditions are critical to solving these equations.

Calculating Position

Position describes an object's location in space. By integrating the velocity function \(v(t)\), we can find the position function \(s(t)\). Here, \(s(t) = -\frac{10}{k} e^{-kt} + \frac{10}{k}\) shows how the position of the object changes over time.

The result indicates that as the velocity decreases, the object's position approaches a certain value, determined by the constants involved. This makes sense conceptually because as the object slows down, it will eventually stop changing its position. The initial condition \(s(0) = 0\) confirms the starting location.

Integrating to find position from velocity is a fundamental concept, and applying initial conditions allows precise determination of movement and final positions.

• Position is derived from integrating the velocity.
• It represents the object's location over time.
• Initial position helps confirm specific solutions.

Role of Initial Conditions

Initial conditions provide the specific starting points for solving problems involving differential equations. In this exercise, both the initial velocity \(v(0) = 10\) and initial position \(s(0) = 0\) play crucial roles in determining the functions that describe an object's motion.

Without initial conditions, the solutions to differential equations involving acceleration, velocity, and position would be incomplete. These conditions are necessary to convert general solutions into specific, real-world applications. They help personalize the differential equations to match exact scenarios, ensuring that predictions and computations reflect the actual behavior of moving objects.

By using initial conditions, mathematicians can confidently solve equations and apply them in fields ranging from physics to engineering.

• Initial conditions set the starting parameters for solving equations.
• They allow for precise calculations and predictions.
• These conditions ensure the solutions are applicable to real-world scenarios.