Question

An object undergoes one-dimensional motion along the \(x\)-axis subject to the given decelerating forces. At time \(t=0\), the object's position is \(x=0\) and its velocity is \(v=v_{0}\). In each case, the decelerating force is a function of the object's position \(x(t)\) or its velocity \(v(t)\) or both. Transform the problem into one having distance \(x\) as the independent variable. Determine the position \(x_{f}\) at which the object comes to rest. (If the object does not come to rest, \(x_{f}=\infty\).) $$ m \frac{d v}{d t}=-k e^{-x} $$

Short answer

Answer: The final position, \(x_f\), of the object when it comes to rest is given by \(x_f = -\frac{m}{k}\ln{v_0}\), where \(m\) is the mass of the object and \(v_0\) is the initial velocity.

Step-by-step solution

Rearrange the equation

Rearrange the given equation to separate variables. We get: $$ m\frac{dv}{-ke^{-x}} = dt $$

Integrate both sides

Integrate both sides of the equation: $$ \int_{v_0}^0 m \frac{dv}{-ke^{-x}}= \int_0^{x_f} dt $$

Substitute 𝑢 = e^(-x)

Now, substitute 𝑢 = e^(-x) into the integral and differentiate 𝑢 with respect to 𝑥: $$ \frac{du}{dx} = -e^{-x} $$ And, $$ dx = -\frac{du}{e^{-x}} $$ Now, the integral equation becomes $$ \int_{v_0}^0 m \frac{dv}{-k u} = \int_{1}^{u_f} -\frac{du}{u} $$ We have to find \(u_f\), which can be related to \(x_f\) since \(u_f = e^{-x_f}\).

Solving the integral

Now, solve the integral $$ -\frac{m}{k}\int_{v_0}^0 \frac{dv}{u} = \int_{1}^{u_f} \frac{du}{u} $$ Then, we have: $$ -\frac{m}{k}\ln{u}\biggr\rvert_{v_0}^0 = \ln{u}\biggr\rvert_1^{u_f} $$

Calculate \(u_f\) and \(x_f\)

Now, calculate \(u_f\): $$ \frac{m}{k}\ln{v_0} = \ln{u_f}-\ln{1} $$ Therefore, $$ u_f = e^{\frac{m}{k}\ln{v_0}} $$ Now, find the final position \(x_f\) using the relation \(u_f = e^{-x_f}\): $$ x_f = -\ln{u_f} = -\frac{m}{k}\ln{v_0} $$ So the final position, \(x_f\) at which the object comes to rest is: $$ x_f = -\frac{m}{k}\ln{v_0} $$

Key concepts

One-Dimensional Motion

Understanding one-dimensional motion is crucial when solving physics problems related to movement along a straight line, such as the exercise at hand. One-dimensional motion refers to the displacement, velocity, and acceleration of an object that is constrained to a single axis, in this case, the x-axis.

In the presented problem, we deal with an object affected by a decelerating force. The initial conditions are given: at time \(t=0\), the position is \(x=0\), with an initial velocity \(v=v_{0}\). The force applied to the object is a function of position \(x(t)\) and velocity \(v(t)\), which affects its velocity and subsequently, its position over time.

To find the final position \(x_{f}\) where the object stops, we need to analyze its motion using the principles of dynamics and the provided differential equation related to the forces acting on the object. In one-dimensional motion problems, it's essential to determine the relationship between time, velocity, and position to solve for the unknown variables.

Separation of Variables

The separation of variables is a mathematical method used to solve ordinary differential equations (ODEs), where we can separate the variables and independently integrate each side of the equation. In our motion problem, the equation governing the motion is an ODE, which can be rearranged so that all terms containing one variable are on one side, and all terms containing the other variable are on the opposite side.

The step-by-step solution begins with rearranging the given ODE to isolate the differential elements \(dv\) and \(dt\) on opposite sides of the equation. This enables us to integrate both sides separately with respect to their own variables. The initial and final conditions are used as limits for the integrals. After completing the integration process, we are left with an expression that can be solved for the unknown, which in this case is the final position \(x_{f}\) where the object comes to rest.

Integrating Factor Method

While the integrating factor method isn't directly used in the solution provided, it's a valuable technique for solving certain types of differential equations, especially linear ODEs, where the standard separation of variables cannot be applied directly.

In problems with differential equations that are not readily separable, an integrating factor is a function that is multiplied by both sides of the ODE to facilitate the separation of variables. The integrating factor is typically a function of one of the variables in the ODE and, once determined, makes the equation easier to solve by turning the left-hand side into the derivative of a product of functions.

Although the integrating factor method is a potent tool in solving differential equations, its application depends on the form of the equation. Since our problem involved a straightforward separation of variables after rearranging, the need for an integrating factor did not arise. However, understanding this method broadens a student's capabilities in tackling a wide array of differential equations.