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Consider an object that is coasting horizontally (positive \(x\) direction) subject to a drag force \(f=-b v-c v^{2} .\) Write down Newton's second law for this object and solve for \(v\) by separating variables. Sketch the behavior of \(v\) as a function of \(t .\) Explain the time dependence for \(t\) large. (Which force term is dominant when \(t\) is large?)

Short Answer

Expert verified
The velocity \( v(t) \) approaches \( -\frac{b}{c} \) as \( t \) becomes large due to the dominance of the \( -cv^2 \) force term.

Step by step solution

01

Write Down Newton's Second Law

Newton's second law states that the sum of the forces on an object equals the object's mass times its acceleration: \[ ma = f \] For this problem, we are given the drag force: \[ f = -b v - c v^2 \] So, Newton's second law becomes: \[ m \frac{dv}{dt} = -b v - c v^2 \] This is the equation we will work with to find velocity \( v \) as a function of time \( t \).
02

Separate Variables

To solve for \( v(t) \), we separate variables in the differential equation:\[ m \frac{dv}{dt} = -b v - c v^2 \] Rearranging terms:\[ \frac{dv}{b v + c v^2} = -\frac{1}{m} dt \] Now, the variables are separated, so we can integrate both sides.
03

Integrate Both Sides

Integrating the left side with respect to \( v \) and the right side with respect to \( t \):\[ \int \frac{dv}{b v + c v^2} = \int -\frac{1}{m} dt \] This requires partial fraction decomposition on the left side, leading to:\[ \frac{1}{b} \ln|b + c v| = -\frac{t}{m} + C \] where \( C \) is the constant of integration.
04

Solve for Velocity \( v(t) \)

Solving the equation from Step 3 for \( v \):\[ b + c v = K e^{-\frac{bt}{m}} \] where \( K \) is an integration constant based on initial conditions. Rearrange for \( v \):\[ v = \frac{K e^{-\frac{bt}{m}} - b}{c} \] This gives \( v(t) \) in terms of \( K \) and the parameters \( b, c, m \).
05

Determine Long-Time Behavior

For large \( t \), the term \( e^{-\frac{bt}{m}} \) approaches zero, so\[ v = -\frac{b}{c} \] This indicates that at large \( t \), the velocity approaches a constant value determined by the parameters \( b \) and \( c \), suggesting the \( c v^2 \) term dominates as the drag becomes significant.
06

Sketch \( v(t) \) and Analyze Dominant Term

Draw a sketch where \( v(t) \) starts from its initial condition and asymptotically approaches \( -\frac{b}{c} \). Initially, \( -bv \) dominates because it is linear, but as \( t \) increases, the \( -cv^2 \) term becomes more significant, enforcing stabilization of \( v \) to a constant value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Drag Force
When an object moves through a fluid or air, it experiences a drag force that opposes the motion. In this exercise, the drag force is represented as \( f = -b v - c v^2 \). This formula combines two types of drag: linear and quadratic.
  • **Linear Drag:** The term \( -b v \) represents linear drag. It is proportional to the velocity \( v \) and comes into play at lower speeds or for streamlined objects.
  • **Quadratic Drag:** The \( -c v^2 \) term denotes quadratic drag. This form of resistance becomes significant at higher speeds, where the object's shape and the density of the fluid highly influence the motion.
The negative signs in the force terms signify that drag acts in the opposite direction to the object's velocity, slowing it down. As the object speeds up, these forces become more prominent, influencing its eventual motion.
Separation of Variables
Separation of variables is a mathematical technique used to solve differential equations, such as Newton's second law in this problem. Here, the equation \( m \frac{dv}{dt} = -b v - c v^2 \) describes the relationship between velocity and time.
To separate variables, we rearrange the equation in a way that one variable is on each side. This looks like: \[ \frac{dv}{b v + c v^2} = -\frac{1}{m} dt \].
This method allows us to tackle the problem by integrating each side independently. The left side is integrated with respect to \( v \) and the right side with respect to \( t \). This process unravels the dynamics of how velocity evolves over time under the influence of drag forces.
Asymptotic Behavior
In examining the asymptotic behavior of a system, we are interested in understanding how the velocity \( v(t) \) behaves as time \( t \) becomes very large. For this specific problem, when \( t \rightarrow \infty \), the exponential term \( e^{-\frac{bt}{m}} \) tends to zero. This leads to the velocity equation simplifying to:
\[ v = -\frac{b}{c} \]
  • This result indicates that the velocity approaches a constant value over time, determined purely by the parameters \( b \) and \( c \).
  • The constant velocity implies that quadratic drag \( -c v^2 \) becomes the dominant force as time progresses, stabilizing the motion.
Understanding this long-term behavior is crucial, as it not only shows the system's stabilization but also highlights the effects of drag in practical applications, such as vehicles moving through air.
Differential Equation Solutions
Solving differential equations often requires finding an explicit expression for the function of interest, in this case, the velocity \( v(t) \). The separated equation \( \frac{dv}{b v + c v^2} = -\frac{1}{m} dt \) needs integration to solve.
Using partial fraction decomposition simplifies the integral of the left side. After integration, we arrive at:
\[ \frac{1}{b} \ln|b + c v| = -\frac{t}{m} + C \], where \( C \) is a constant determined by initial conditions.
Solving for \( v \), we get:
\[ v(t) = \frac{K e^{-\frac{bt}{m}} - b}{c} \]
Here, \( K \) integrates initial conditions, ensuring the solution is specific to the scenario's starting details. Understanding the procedure for solving these equations equips students to address a variety of problems involving changing rates, which are common in both theoretical and practical physics contexts.

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Most popular questions from this chapter

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