Problem 48 - Master Resource Book in JEE Main Physics

Question

A particle located at \(x=0\) at time \(t=0\), starts moving along the positive \(x\)-direction with a velocity \(v\) that varies as \(v=a \sqrt{x}\). The displacement of the particle varies with time as (a) \(t^{3}\) (b) \(t^{2}\) (c) \(t\) (d) \(t^{1 / 2}\)

Step-by-Step Solution

Answer

Displacement varies as \( t^2 \); correct answer is (b) \( t^2 \).

Step 1: Understanding the Problem

We need to find how displacement \( x \) of a particle changes with time \( t \) given that its velocity \( v = a \sqrt{x} \). This means we need to express \( x \) as a function of \( t \).

Step 2: Relating Velocity to Time and Displacement

Velocity is the rate of change of displacement over time, so we can write \( v = \frac{dx}{dt} = a \sqrt{x} \). Rearrange this equation: \( dt = \frac{dx}{a \sqrt{x}} \).

Step 3: Integrating to Find the Time-Displacement Relationship

Integrate both sides to find the relationship between \( x \) and \( t \). The left side integrates with respect to \( t \) and the right with respect to \( x \): \[ \int dt = \int \frac{dx}{a \sqrt{x}} \]The integration gives \( t = \frac{2}{a} \sqrt{x} + C \).

Step 4: Using Initial Conditions to Find the Constant of Integration

Since at \( t = 0 \), \( x = 0 \), plug these into the equation \( t = \frac{2}{a} \sqrt{x} + C \) to find \( C \). This gives \( C = 0 \), so \( t = \frac{2}{a} \sqrt{x} \).

Step 5: Expressing Displacement as a Function of Time

Solve the equation \( t = \frac{2}{a} \sqrt{x} \) for \( x \). Rearrange to get \( \sqrt{x} = \frac{a}{2} t \), then \( x = \left( \frac{a}{2} t \right)^2 = \frac{a^2}{4} t^2 \). This shows displacement \( x \) is proportional to \( t^2 \).

Step 6: Select Correct Answer

The derived relationship shows the displacement \( x \) varies with the square of time \( t^2 \). So, the correct answer is (b) \( t^2 \).

Key Concepts

Velocity and Displacement Relationship

Understanding the relationship between velocity and displacement is key to comprehending motion in kinematics. Velocity denotes the rate at which an object's position changes with time. In mathematical terms, it is defined as the derivative of displacement \( x \) with respect to time \( t \), represented by \( v = \frac{dx}{dt} \). In our problem, the velocity \( v \) is given as a function of displacement, specifically \( v = a \sqrt{x} \). This connects how the velocity changes depending on the particle's current position, or displacement. Understanding this allows us to explore how changes in velocity translate to changes in displacement. To decipher how \( x \) evolves over time, we use integration as a tool to reverse the differentiation represented by velocity.

Integration in Kinematics

Integration is a powerful technique in kinematics, often used to transition from rates of change, like velocity, to accumulated quantities, such as displacement. Knowing velocity as \( v = a \sqrt{x} \), we set up the equation \( v = \frac{dx}{dt} = a \sqrt{x} \) to relate velocity with displacement over time. Rearranging gives \( dt = \frac{dx}{a \sqrt{x}} \). This separation allows for integration of both sides to establish a relationship between \( x \) and \( t \). The integration yields: \[ \int dt = \int \frac{dx}{a \sqrt{x}} \]After integrating, we derive the equation for time in terms of displacement: \( t = \frac{2}{a} \sqrt{x} + C \). Here, \( C \) is the constant of integration, representing initial conditions which can be determined with additional information. Through this integration, we convert a complex relationship into a more manageable form that shows displacement's dependence on time.

Initial Conditions in Motion Problems

In solving motion problems, initial conditions are critical to find specific solutions for variables like displacement or time. They are essential in determining unknown constants that arise during integration. In this exercise, the initial condition is given as \( x = 0 \) when \( t = 0 \). This means the particle starts from a specific point, which allows us to solve for the constant \( C \) in the integrated equation. Substituting these initial values into \( t = \frac{2}{a} \sqrt{x} + C \) simplifies to \( 0 = \frac{2}{a} \sqrt{0} + C \). This directly shows that \( C = 0 \). Incorporating these initial conditions gives us a precise equation: \( t = \frac{2}{a} \sqrt{x} \). Resolving for \( x \), we find that displacement grows proportional to time squared, specifically \( x = \left( \frac{a}{2} t \right)^2 = \frac{a^2}{4} t^2 \). Thus, initial conditions ensure our solution accurately reflects the specific scenario described.