Problem 16 - Master Resource Book in JEE Main Physics

Question

The motion of a body is given by the equation \(\frac{d v(t)}{d t}=6.0-3 v(t)\), where \(v(t)\) is speed in \(\mathrm{ms}^{-1}\) and \(t\) in second. If body was at rest at \(t=0\) (a) the terminal speed is \(2.0 \mathrm{~ms}^{-1}\) (b) the speed varies with the times as \(v(t)=2\left(1-e^{-3 t}\right) \mathrm{ms}^{-1}\) (c) the speed is \(1.0 \mathrm{~ms}^{-1}\) when the acceleration is half the initial value (d) the magnitude of the initial acceleration is \(6.0 \mathrm{~ms}^{-2}\)

Step-by-Step Solution

Answer

All options (a, b, c, d) are correct.

Step 1: Analyze Differential Equation

The motion is described by the differential equation \( \frac{dv(t)}{dt} = 6.0 - 3v(t) \). This is a first-order linear differential equation with variables separable.

Step 2: Separate Variables and Integrate

Rearrange the equation to separate variables: \( \frac{1}{6.0 - 3v(t)}dv = dt \). Integrate both sides to solve for \(v(t)\): \( \int \frac{1}{6.0 - 3v} dv = \int dt \).

Step 3: Solve the Integrals

The integral on the left is \(-\frac{1}{3} \ln|6.0 - 3v|\), and the right side is \(t + C\). Solving gives: \( -\ln|6.0 - 3v| = 3t + C' \).

Step 4: Solve for \(v(t)\)

Exponentiate both sides: \(6.0 - 3v = Ce^{-3t}\) where \(C = e^{C'}\). By initial condition \(v(0) = 0\), we find \(C = 6\). So \(v(t) = 2(1 - e^{-3t})\).

Step 5: Find Terminal Speed

As \(t \to \infty\), \(e^{-3t} \to 0\), so the terminal speed is \(v = 2.0 \text{ ms}^{-1}\).

Step 6: Initial Acceleration

Differentiate \(v(t)\) to find acceleration \(a(t) = \frac{dv}{dt} = 6.0 - 3v(t)\). At \(t = 0\), \(v(0) = 0\), so \(a(0) = 6.0 \text{ ms}^{-2}\).

Step 7: Find Speed when Acceleration is Half Initial

Set \(a(t) = 3.0\), solving \(6.0 - 3v = 3.0\), gives \(v = 1.0 \text{ ms}^{-1}\).

Key Concepts

Understanding Linear Differential Equations

To understand the motion of a body in physics, we often use differential equations. A linear differential equation is one such type, characterized by its form which involves derivatives of a function. In the context of our problem, the equation is \[ \frac{d v(t)}{d t} = 6.0 - 3 v(t) \] This equation is linear because the terms involving the derivative and the function \(v(t)\) do not multiply each other or involve any powers besides the first degree. This structure makes them particularly friendly to solve as they obey the principle of superposition.
In linear differential equations like this one, solving involves finding the function \(v(t)\) that fits. Such equations frequently appear in physics problems, helping us model real-world phenomena like the motion of objects, electronic circuits, or heat transfer.

Exploring Terminal Velocity

Terminal velocity refers to the constant speed an object reaches when the forces acting upon it balance out. At this point, acceleration ceases because the driving force and resistive forces are equal. In our example, the body reaches a terminal velocity when \[ 6.0 - 3v(t) = 0 \] Solving this, we find \[ v(t) = 2.0 \text{ ms}^{-1} \] This is the speed at which the net force on the object is zero. It means that the consistent input of energy is perfectly counterbalanced by resistive forces, such as friction or drag.
Understanding terminal velocity helps us analyze how objects like parachutes or falling raindrops behave. It is the point where they stop accelerating and move steadily until they meet a new influencing factor.

Grasping Initial Acceleration

Initial acceleration is simply the rate of change of velocity when time \(t = 0\). Calculating it gives us insight into how quickly a body begins to move from rest. This is crucial for understanding the kinetics of an object.In our scenario, this is derived by differentiating the velocity function again: \[ a(t) = \frac{dv}{dt} = 6.0 - 3v(t) \]At time \(t = 0\), with the initial velocity \(v(0) = 0\), the initial acceleration equals \[ a(0) = 6.0 \text{ ms}^{-2} \] This value tells us the rate at which the object's velocity will increase from zero, helping us predict its motion at the start.
Initial acceleration is fundamental for trajectories, determining how fast something like a car or a projectile will start moving from rest.

The Variable Separation Method Explained

One of the powerful techniques to solve differential equations, like in our problem, is the variable separation method. It shines with equations where you can isolate the variables to opposite sides of the equation. For the differential equation \[ \frac{d v(t)}{d t} = 6.0 - 3 v(t) \] We separate variables to solve: \[ \frac{1}{6.0 - 3v(t)} dv = dt \] This allows us to integrate both sides individually, leading to a solution that expresses \(v(t)\) explicitly in terms of \(t\).
The variable separation method is intuitive and widely applicable; it allows you to straightforwardly integrate and obtain solutions. This method is particularly handy in physics, where it’s often the cleanest way to tackle equations describing natural phenomena.