Problem 16 - Master Resource Book in JEE Main Physics

Question

The retardation experienced by a moving motor boat, after its engine is cut- off, is given by \(\frac{d v}{d t}=-k v^{3}\), where \(k\) is a constant. If \(v_{0}\) is the magnitude of the velocity at cut-off, the magnitude of the velocity at time \(t\) after the cut-off is (a) \(v_{0}\) (b) \(\frac{v_{0}}{2}\) (c) \(v_{0} e^{-k t}\) (d) \(\frac{v_{0}}{\sqrt{2 v_{0}^{2} k t+1}}\)

Step-by-Step Solution

Answer

The velocity at time \( t \) is \( \frac{v_0}{\sqrt{2 v_0^{2} k t + 1}} \).

Step 1: Differential Equation Setup

The problem provides a relationship between velocity \( v \) and time \( t \) through the differential equation \( \frac{d v}{d t} = -k v^{3} \). Our task is to find \( v \) as a function of \( t \).

Step 2: Separate Variables

Separate the variables to integrate: \[ \frac{d v}{v^{3}} = -k \cdot dt \].

Step 3: Integrate Both Sides

Integrate both sides of the equation. The left side becomes \[ \int \frac{d v}{v^{3}} = \int -k \cdot dt \].The integration results in \[ -\frac{1}{2 v^{2}} = -k t + C \], where \( C \) is a constant of integration.

Step 4: Solve for Integration Constant

Apply initial conditions to solve for \( C \). We know at \( t = 0, v = v_0 \). Substitute these into the integrated equation: \[ -\frac{1}{2 v_0^{2}} = C \].

Step 5: Substitute Back the Constant

Substitute \( C \) back into the expression: \[ -\frac{1}{2 v^{2}} = -k t -\frac{1}{2 v_0^{2}} \].

Step 6: Solve for v

Rearrange to find \( v \): \[ \frac{1}{2 v^{2}} = k t + \frac{1}{2 v_0^{2}} \] which simplifies to: \[ v^{2} = \frac{1}{2(k t v_0^{2} + \frac{1}{2})} \].Taking the square root gives:\[ v = \frac{1}{\sqrt{2 v_0^{2} k t + 1}} \].

Key Concepts

Velocity-Time Relationship

The velocity-time relationship is crucial to understanding how objects move over time. In physics, velocity describes how fast an object changes its position. It is often represented as a function of time, illustrating how the speed varies as time progresses. In the context of our problem, the velocity of a motorboat influences how its motion slows down over time, especially when the engine is cut-off. The provided differential equation \( \frac{d v}{d t} = -k v^{3} \) is a mathematical representation that indicates how the velocity changes with the passage of time.

• The function \( v(t) \) describes velocity as it changes over time \( t \).
• The negative sign in front of the equation implies a decrease in velocity, indicative of retardation or deceleration.
• Understanding this relationship is key to solving the differential equation and predicting future velocities of the object.

Retardation in Motion

Retardation, or deceleration, refers to the reduction of velocity with respect to time. In our exercise, retardation occurs because the motorboat is slowing down after the engine is stopped. The rate of this slowdown is directly affected by its current velocity and is represented by the equation \( \frac{d v}{d t} = -k v^{3} \).Let's delve into what this equation tells us:

• The cube of the velocity term \( v^3\) suggests that speed decreases rapidly when velocity is high, and less rapidly as velocity diminishes.


• The constant \( k \) governs the rate of retardation. Larger values of \( k \) indicate a quicker slowdown.


• This equation provides an intuitive way to model situations in which friction or resistance depends more dramatically on speed.

This understanding can also apply to other real-life scenarios where retardation plays a role, such as brakes slowing down a car or air resistance affecting a falling object's speed.

Integration in Calculus

Integration in calculus is a fundamental method used to find functions from their rates of change. When dealing with differential equations, integration helps unravel how a function, such as velocity, changes over time.In our problem, we used integration to calculate the velocity of the motorboat after the engine is stopped. The differential equation \( \frac{d v}{d t} = -k v^{3} \) was separated into an integrable form, and then both sides were integrated:

• The left side \( \int \frac{d v}{v^{3}} \) involved integrating with respect to velocity, revealing how the velocity depends on initial conditions.


• The right side involved a simple integration \( \int -k \cdot dt \) over time, contributing to understanding how long the retardation lasts.

After finding the integral, the constants of integration were determined by applying given initial conditions, specifically at \( t = 0 \) and \( v = v_{0} \). This process is quintessential in physics as it allows deriving exact solutions that follow initial physical conditions imposed on the system.