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Magazine Chapter 3: Problem 50
A point moves in a straight line under the retardation \(k v^{2}\). If the initial velocity is \(u\), the distance covered in \(t\) second is : (a) kut (b) \(\frac{1}{k} \log k u t\) (c) \(\frac{1}{k} \log (1+k u t)\) (d) \(k \log k u t\)
Short Answer
Expert verified
The correct answer is (c) \(\frac{1}{k} \log(1+k u t)\).
Step by step solution
01
Understand the Problem
We need to determine the distance covered by a point moving under a retardation proportional to the square of its velocity. The given retardation is \( kv^2 \), and we need to find a formula for distance using this information with an initial velocity of \( u \).
02
Establish the Differential Equation
The retardation \( kv^2 \) implies the acceleration is \( -kv^2 \). Thus, we have the differential equation: \[ \frac{dv}{dt} = -kv^2 \]
03
Separate Variables and Integrate
To solve \( \frac{dv}{dt} = -kv^2 \), separate the variables: \[ \int \frac{1}{v^2} \, dv = -k \int dt \]Integrate both sides: \[ -\frac{1}{v} = -kt + C \]
04
Solve for Constant of Integration
Using the initial condition, when \( t = 0 \), \( v = u \):\[ -\frac{1}{u} = C \]Thus, the equation becomes:\[ -\frac{1}{v} = -kt - \frac{1}{u} \]
05
Rearrange for Velocity
Rearrange the equation for \( v \):\[ \frac{1}{v} = kt + \frac{1}{u} \]\[ v = \frac{1}{kt + \frac{1}{u}} \]
06
Relate Velocity to Distance
Since velocity is derivative of position, \( v = \frac{ds}{dt} \):\[ \frac{ds}{dt} = \frac{1}{kt + \frac{1}{u}} \]Separate variables and integrate:\[ ds = \frac{1}{kt + \frac{1}{u}} \, dt \]
07
Integrate to Find Distance
Integrate both sides:\[ s = \int \frac{1}{kt + \frac{1}{u}} \, dt = \frac{1}{k} \log(kt + \frac{1}{u}) + C' \]Using initial condition \( s = 0 \) at \( t = 0 \): \( C' = -\frac{1}{k} \log\left(\frac{1}{u}\right) \).So,\[ s = \frac{1}{k} \log\left((kt + \frac{1}{u})u\right) = \frac{1}{k} \log(1 + kut) \]
08
Compare to Given Options
The distance matches option (c): \( \frac{1}{k} \log(1 + kut) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations play a crucial role in understanding motion, particularly when there's a change in velocity or acceleration. In the given exercise, the motion is influenced by a retardation, which is essentially negative acceleration, proportional to the square of the velocity. This is expressed as the differential equation \( \frac{dv}{dt} = -kv^2 \), where \( v \) is velocity, \( t \) is time, and \( k \) is a constant of proportionality.
Such equations describe how one quantity changes with another. Here, it tells us how velocity changes over time. By solving these equations, we can derive various functions such as velocity-time or distance-time relationships.
Such equations describe how one quantity changes with another. Here, it tells us how velocity changes over time. By solving these equations, we can derive various functions such as velocity-time or distance-time relationships.
- **Separation of Variables**: This technique is commonly used to solve differential equations by isolating \( dv \) and \( dt \). This allows us to integrate both sides with respect to their variables.
- **Integration**: Once variables are separated, integrating helps find solutions that represent velocity or position as functions of time.
Initial Velocity
Initial velocity refers to the speed at which an object starts moving. In many physics problems, it's a critical value, as it often acts as a boundary or initial condition when solving equations. Here, the initial velocity is given as \( u \).
When we solve the differential equation, knowing the initial velocity allows us to determine the constant of integration. At \( t = 0 \), the velocity \( v \) is equal to \( u \), leading to the equation: \[-\frac{1}{u} = C\] This boundary condition is vital as it allows us to accurately define functions that model the system, ensuring any subsequent derived values, like distance, are accurate.
When we solve the differential equation, knowing the initial velocity allows us to determine the constant of integration. At \( t = 0 \), the velocity \( v \) is equal to \( u \), leading to the equation: \[-\frac{1}{u} = C\] This boundary condition is vital as it allows us to accurately define functions that model the system, ensuring any subsequent derived values, like distance, are accurate.
- **Importance**: Determines the constant of integration, enabling precise calculations.
- **Application**: Used as the starting point for modeling any further motion.
Velocity-Time Relationship
Understanding the velocity-time relationship is essential in analyzing motion. Velocity is the rate of change of position with respect to time. In the presence of retardation, such as \( -kv^2 \), the velocity decreases over time according to the relationship we derived earlier.
From the solution, after rearranging and integrating the differential equation, we established:\[ v = \frac{1}{kt + \frac{1}{u}} \]This equation illustrates how velocity changes inversely with time under constant retardation. It declines as time increases, reflecting the object's slowing down.
From the solution, after rearranging and integrating the differential equation, we established:\[ v = \frac{1}{kt + \frac{1}{u}} \]This equation illustrates how velocity changes inversely with time under constant retardation. It declines as time increases, reflecting the object's slowing down.
- **Interpretation**: Shows the deceleration of the object, useful in deriving the distance-time relationship.
- **Significance**: Effective for predicting future positions by transitioning from velocity to distance.
Integration in Physics
Integration is a powerful mathematical tool in physics for determining quantities like displacement from velocities. Once we have an expression for velocity as a function of time, like \( v = \frac{1}{kt + \frac{1}{u}} \), integrating can help us find the distance or displacement of the object.
The process involves evaluating the integral of velocity with respect to time, which gives:\[ s = \frac{1}{k} \log(1 + kut) \]This result comes from integrating the separated equation:\[ \frac{ds}{dt} = \frac{1}{kt + \frac{1}{u}} \]Integration allows us to understand how the object moves over time by knowing its velocity behavior.
The process involves evaluating the integral of velocity with respect to time, which gives:\[ s = \frac{1}{k} \log(1 + kut) \]This result comes from integrating the separated equation:\[ \frac{ds}{dt} = \frac{1}{kt + \frac{1}{u}} \]Integration allows us to understand how the object moves over time by knowing its velocity behavior.
- **Utility**: Translates changes in velocity into spatial terms such as distance.
- **Application**: Used across various domains in physics to solve problems involving motion and rates of change.
