Problem 49 - Master Resource Book in JEE Main Physics

Question

An object, moving with a speed of \(6.25 \mathrm{~m} / \mathrm{s}\), is declerated at a rate given by \(\frac{d v}{d t}=-2.5 \sqrt{v}\), where \(v\) is the instantaneous speed. The time taken by the object, to come to rest would be (a) \(2 \mathrm{~s}\) (b) \(4 \mathrm{~s}\) (c) \(8 \mathrm{~s}\) (d) \(1 \mathrm{~s}\)

Step-by-Step Solution

Answer

The object takes 2 seconds to come to rest.

Step 1: Understand the Problem

The object decelerates with an equation \( \frac{dv}{dt} = -2.5\sqrt{v} \). We start with an initial speed \( v = 6.25 \) m/s and want to find the time taken for the speed to become zero.

Step 2: Set Up the Differential Equation

The differential equation given is \( \frac{dv}{dt} = -2.5\sqrt{v} \). To solve for time, we need to integrate this equation.

Step 3: Separate the Variables

Re-arrange the equation to separate variables: \( \frac{dv}{\sqrt{v}} = -2.5 \, dt \). This allows us to integrate both sides separately.

Step 4: Integrate Both Sides

Integrate the left-hand side \( \int \frac{dv}{\sqrt{v}} \) which equals \( 2\sqrt{v} \). Integrate the right-hand side \( \int -2.5 \, dt \) which equals \( -2.5t + C \).

Step 5: Solve the Integral Equation

After integration, we have \( 2\sqrt{v} = -2.5t + C \). Substitute \( v = 6.25 \) m/s for when \( t = 0 \) to find \( C \).

Step 6: Determine Constant of Integration

Substitute \( t = 0, \, v = 6.25 \) into \( 2\sqrt{6.25} = -2.5 \times 0 + C \), which simplifies to \( 2\times 2.5 = C \) giving \( C = 5 \).

Step 7: Solve for Time When the Object Stops

Substitute \( v = 0 \) and solve for \( t \): \( 0 = -2.5t + 5 \). Thus, \( 2.5t = 5 \) and \( t = \frac{5}{2.5} = 2 \).

Key Concepts

Differential Equations

A differential equation is a type of equation that involves derivatives, which can show how a quantity changes over time. In physics, they are essential for modeling dynamic systems.

For a moving object that is slowing down, such as in our exercise, the change in speed (\(v\)) is represented by the differential equation \(\frac{dv}{dt} = -2.5\sqrt{v}\). Here, \(\frac{dv}{dt}\) is the derivative of the speed with respect to time, indicating how speed changes at any moment. This negative sign ahead of the terms tells us that the speed is decreasing, a phenomenon we can describe as deceleration -- which we'll delve into next.

These equations can be solved to predict future states of the system, such as finding out when the object will stop moving. They serve as powerful tools to set the groundwork for analysis and integration in mathematical physics.

Deceleration

Deceleration is simply the process of an object slowing down, i.e., a negative acceleration. It is an essential concept in kinematics.

Deceleration is expressed in terms of the rate of change of velocity. In the exercise, deceleration is modeled by the term \(-2.5\sqrt{v}\), where \(v\) is the instantaneous speed. The function shows that the speed decreases more rapidly when it is higher, and more gently as the speed approaches zero.

• The negative sign is crucial as it indicates that the velocity is decreasing.
• The term \(\sqrt{v}\) suggests that as the speed reduces, the deceleration rate slows down too.

Understanding deceleration helps in solving problems related to stopping times and distances, making it a fundamental part of kinematics.

Instantaneous Speed

Instantaneous speed refers to the speed of an object at any given moment in time. It is a snapshot of an object's velocity at precisely one point, rather than over an interval. Unlike average speed, instantaneous speed does not consider the total distance or time traveled, just the immediate rate of motion.

In the given exercise, we start with an initial instantaneous speed of 6.25 m/s, which indicates how fast the object is moving right when we begin observing it. As time progresses and deceleration occurs as described by the differential equation, this instantaneous speed decreases until the object comes to a stop.

• This value of 6.25 m/s becomes an initial condition for our calculations.
• The relationship between speed and deceleration explains how the speed changes over time according to the given differential equation.

Instantaneous speed can help us understand how an object's motion evolves during kinematic changes.

Integration

Integration is a core mathematical process used to solve differential equations, convert rate-based information into situational state, and find accumulated change over time. It's the reverse operation of differentiation.

In our example, we perform integration to solve the differential equation \(\frac{dv}{dt} = -2.5\sqrt{v}\).

• Separating variables, we rewrite it as \(\frac{dv}{\sqrt{v}} = -2.5 \, dt\)
• We then integrate both sides. The left-hand side results in \(2\sqrt{v}\)
• The right-hand side integrates to \(-2.5t + C\), where C is a constant of integration.

Integration not only lets us determine the functional relationship between variables but also helps solve for unknowns, such as time to stop in this exercise. Determining the constant of integration is a step involving initial conditions, such as knowing the speed at a certain time, which ensures the solution corresponds accurately to the specific scenario at hand.