Problem 71 - Master Resource Book in JEE Main Physics

Question

The coordinates of a moving particle at time \(t\) are given by \(x=c t^{2}\) and \(y=b t^{2}\). The instantaneous speed of the particle is (a) \(2 t(b+c)\) (b) \(2 t(b+c)^{1 / 2}\) (c) \(2 t\left(c^{2}-b^{2}\right)\) (d) \(2 t\left(c^{2}+b^{2}\right)^{1 / 2}\)

Step-by-Step Solution

Answer

The instantaneous speed of the particle is option (d): \(2 t\left(c^{2}+b^{2}\right)^{1 / 2}\).

Step 1: Differentiate x and y with respect to time t

First, we need to find the velocity components of the particle in both the x and y directions. For this, differentiate the given equations for x and y with respect to time.For the x-coordinate: \[ x = c t^2 \]\[ \frac{dx}{dt} = \frac{d}{dt}(c t^2) = 2ct \]For the y-coordinate: \[ y = b t^2 \]\[ \frac{dy}{dt} = \frac{d}{dt}(b t^2) = 2bt \]

Step 2: Use Pythagorean Theorem to Find Instantaneous Speed

To find the instantaneous speed of the particle, we use the Pythagorean theorem to combine the velocities in the x and y directions.The instantaneous speed \( v \) is given by:\[ v = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \]Substitute the expressions for \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) we found in Step 1:\[ v = \sqrt{(2ct)^2 + (2bt)^2} \]

Step 3: Simplify the Expression for Instantaneous Speed

Simplify the expression obtained for the instantaneous speed.\[ v = \sqrt{4c^2t^2 + 4b^2t^2} \]Factor out \(4t^2\) from the square root:\[ v = \sqrt{4t^2(c^2 + b^2)} \]Take the square root of each part of the expression:\[ v = 2t\sqrt{c^2 + b^2} \]

Step 4: Identify the Correct Answer

Compare the simplified expression for the instantaneous speed with the given options:The expression \( v = 2t\sqrt{c^2 + b^2} \) corresponds to option (d): \(2 t\left(c^{2}+b^{2}\right)^{1 / 2}\).

Key Concepts

Differentiation in Physics

Differentiation in physics is vital for understanding how quantities change over time. It helps in analyzing motion patterns by focusing on how position variables evolve. In this context, we differentiate the position functions for a moving particle to derive the velocity components. This process involves applying the basic differentiation rules to the given position functions. For example, consider a particle whose coordinates are given by the equations \(x = c t^2\) and \(y = b t^2\). By differentiating \(x\) with respect to time \(t\), we find its rate of change, known as the velocity in the x-direction, resulting in \(\frac{dx}{dt} = 2ct\). Similarly, differentiating \(y\) provides the rate of change in the y-direction, \(\frac{dy}{dt} = 2bt\).The derivatives essentially describe how fast the x and y coordinates change as time progresses, representing the particle's velocity components. Understanding differentiation in physics lets us analyze complex motion scenarios more simply and find quantities like instantaneous speed effectively.

Velocity Components

Velocity components are fundamental to breaking down and analyzing the motion of objects in different directions. When a particle moves in two dimensions, it's helpful to analyze its motion along each individual axis first.The velocity components are determined by differentiating the position functions. In our example:- The x-component of velocity is \(v_x = \frac{dx}{dt} = 2ct\).- The y-component of velocity is \(v_y = \frac{dy}{dt} = 2bt\).These components describe how fast the particle is moving along the x-axis and the y-axis, respectively. These values form the building blocks to find the total velocity magnitude, also referred to as instantaneous speed.Understanding the individual velocity components can help in predicting the particle's future position. It also assists in solving for other motion-related parameters when examined within a given time frame. This method is a practical application of breaking down a complex problem into more manageable parts.

Pythagorean Theorem in Physics

The Pythagorean theorem plays a crucial role in physics, particularly when calculating resultant quantities in two-dimensional motion. It allows us to combine orthogonal components, like velocity in the x and y directions, to find a total value, often referred to as magnitude.In our exercise, the Pythagorean theorem combines velocity components to determine the particle's instantaneous speed. Using the formula:\[ v = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \]Substituting the known derivatives, we have:\[ v = \sqrt{(2ct)^2 + (2bt)^2} \]Further simplifying involves factoring out common terms and simplifying the expression:\[ v = \sqrt{4t^2(c^2 + b^2)} = 2t\sqrt{c^2 + b^2} \]This approach exemplifies how the Pythagorean theorem is applied in physics to link individual directional components together, helping to describe the overall magnitude, like speed, in a comprehensive and efficient manner. It provides a foundation for interpreting results accurately when analyzing two-dimensional motion.