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A test projectile is fired horizontally into a viscous liquid with a velocity \(v_{0} .\) The retarding force is proportional to the square of the velocity, so that the acceleration becomes \(a=-k v^{2}\). Derive expressions for the distance \(D\) traveled in the liquid and the corresponding time \(t\) required to reduce the velocity to \(v_{0} / 2 .\) Neglect any vertical motion.

Short Answer

Expert verified
Time \(t = \frac{1}{kv_0}\), and distance \(D = \frac{1}{k} \ln(2)\).

Step by step solution

01

Understanding the Problem

We have a projectile entering a viscous liquid with an initial horizontal velocity \(v_0\). The deceleration due to the liquid is given by \(a = -k v^2\), where \(k\) is a constant, and our task is to find the distance traveled \(D\) and the time \(t\) it takes for the velocity to reduce to \(v_0/2\).
02

Set up the Differential Equation for Velocity

Start with the formula for acceleration: \(a = \frac{dv}{dt} = -k v^2\). Rearrange it to get the differential equation \(\frac{dv}{v^2} = -k \, dt\).
03

Integrate the Velocity Function

Integrate both sides to find the velocity as a function of time: \(\int \frac{dv}{v^2} = -\int k \, dt\). This gives \(-\frac{1}{v} = -kt + C\), where \(C\) is the integration constant.
04

Solve for the Constant of Integration

Apply the initial condition \(v = v_0\) when \(t = 0\) to find the constant \(C\): \(-\frac{1}{v_0} = C\). Substitute \(C\) back into the equation to get \(-\frac{1}{v} = -kt - \frac{1}{v_0}\).
05

Find Time to Reduce Velocity to Half

Solve \(-\frac{1}{v} = -kt - \frac{1}{v_0}\) for \(t\) when \(v = \frac{v_0}{2}\): \(\frac{1}{v_0/2} = kt + \frac{1}{v_0}\). Simplifying gives \(kt = \frac{2}{v_0} - \frac{1}{v_0}\), so \(t = \frac{1}{kv_0}\).
06

Set up the Integral for Distance

Write velocity as a function of distance: \(v = \frac{dx}{dt}\). Use \(dx = v \, dt\), and substitute the integrated velocity \(v = \frac{1}{kt + \frac{1}{v_0}}\) into this equation to get \(dx = \frac{1}{kt + \frac{1}{v_0}} \, dt\).
07

Integrate to Find Distance

Integrate both sides \(\int dx = \int \frac{1}{kt + \frac{1}{v_0}} \, dt\). Simplifying, we find \(D = \ln\left(\frac{2}{v_0}\frac{1}{v_0}\)\) which simplifies via the constant multiple rule for \(\ logarithms to: \(\frac{1}{k} \ln\left(\frac{v0}{v}\right)\).
08

Calculate Distance Specifically

Substitute \(v = \frac{v_0}{2}\) into the expression, the distance \(D = \frac{1}{k} \ln(2)\) which relates the distance traveled once the velocity halved.

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Key Concepts

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

Viscous Damping
Viscous damping is an essential concept in physics that describes the resistance enacted by a fluid, like a liquid, against the motion of an object passing through it. When a projectile is fired into a viscous liquid, as in the case of our exercise, the liquid's viscosity exerts a drag force that opposes the projectile's motion.
This damping force is often proportional to the square of the velocity of the projectile. In mathematical terms, if the velocity of the object is denoted as \(v\), the viscous damping force \(F_d\) can be expressed as \(F_d = -k v^2\), where \(k\) is a constant that depends on the properties of the fluid and the object.
Understanding viscous damping helps us to predict how quickly different fluids can slow down a moving object. It is not only critical in engineering and mechanics but also in various applications ranging from automobiles to biological systems.
Differential Equations
Differential equations are equations that involve the rates at which quantities change. They are indispensable in describing the behavior of dynamic systems, such as the deceleration of a projectile in a viscous fluid.
In the exercise, we use a differential equation to relate the acceleration and velocity of the projectile: \(a = -k v^2\). By expressing acceleration as the derivative of velocity with respect to time, \(a = \frac{dv}{dt}\), we obtain the equation \(\frac{dv}{dt} = -k v^2\).
Solving this differential equation requires integrating both sides, which allows us to find how velocity changes over time. Understanding these equations helps us to predict how quickly an object will slow down, given an initial speed and the parameters of the resistive force.
Retarding Force
A retarding force is any force that opposes the motion of an object, effectively slowing it down, and in the context of our exercise, that force is due to viscous damping.
This force is mathematically related to the velocity of the projectile, specifically given by \(F_r = -k v^2\). Because it opposes motion, the retarding force is negative, acting opposite to the direction of the projectile's initial velocity.
The significance of understanding retarding forces lies in their impact on the object’s kinetic energy. In our projectile example, as the retarding force acts, it continually reduces the velocity, which ultimately affects both how far the projectile travels and how long it takes to reach certain points before stopping.
Velocity Reduction
Velocity reduction is a key topic when analyzing the motion of objects under resistive forces, such as viscous damping. In our exercise, the projectile's initial velocity \(v_0\) is halved as it moves through the fluid.
The process of solving for the distance \(D\) and time \(t\) taken for the velocity to decrease to \(\frac{v_0}{2}\) involves integrating the relationship \(\frac{dv}{v^2} = -k \, dt\), yielding insights into how resistance affects motion over time.
Practically, velocity reduction is crucial in designing systems where controlled deceleration is required, such as in safety systems of vehicles or in the design of sports equipment. Understanding how this reduction occurs allows engineers to make better-informed decisions about material choices and system designs.

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