Problem 9 - Master Resource Book in JEE Main Physics

Question

A particle moves in the \(x y\)-plane with velocity \(v_{x}=8 t-2\) and \(v_{y}=2\) If it passes through the point \(x=14\) and \(y=4\) at \(t=2\) s, find the equation \((x-y\) relation) of the path. (a) \(x=y^{2}-y+2\) (b) \(x=2 y^{2}+2 y-3\) (c) \(x=3 y^{2}+5\) (d) Cannot be found from above data

Step-by-Step Solution

Answer

(a) \(x = y^2 - y + 2\) is the correct equation for the path.

Step 1: Find Position Functions

To find the path of the particle, we need to integrate the velocity functions to find the position functions. The velocity functions are given as \(v_x = 8t - 2\) and \(v_y = 2\). Integrating \(v_x\) with respect to time gives the position function for \(x\), while integrating \(v_y\) gives the position function for \(y\).

Step 2: Integrate Velocity Function for x

The velocity function for \(x\) is \(v_x = 8t - 2\). Integrate this with respect to time: \[ x(t) = \int (8t - 2) \, dt = 4t^2 - 2t + C_1 \] where \(C_1\) is the constant of integration.

Step 3: Determine Constant for x

Use the initial condition \(x = 14\) at \(t = 2\) to find \(C_1\). Substituting these values into the position function: \[ 14 = 4(2)^2 - 2(2) + C_1 \]\[ 14 = 16 - 4 + C_1 \]\[ 14 = 12 + C_1 \]\[ C_1 = 2 \]So the position function for \(x\) becomes \(x(t) = 4t^2 - 2t + 2\).

Step 4: Integrate Velocity Function for y

The velocity function for \(y\) is \(v_y = 2\). Integrate this with respect to time: \[ y(t) = \int 2 \, dt = 2t + C_2 \] where \(C_2\) is the constant of integration.

Step 5: Determine Constant for y

Use the initial condition \(y = 4\) at \(t = 2\) to find \(C_2\). Substituting these values into the position function: \[ 4 = 2(2) + C_2 \]\[ 4 = 4 + C_2 \]\[ C_2 = 0 \]So the position function for \(y\) becomes \(y(t) = 2t\).

Step 6: Eliminating Parameter t

To find the equation relating \(x\) and \(y\), eliminate the parameter \(t\) from the equations \(x(t) = 4t^2 - 2t + 2\) and \(y(t) = 2t\). From \(y = 2t\), express \(t\) as \(t = \frac{y}{2}\). Substitute \(t\) in the equation of \(x(t)\):\[ x = 4\left( \frac{y}{2} \right)^2 - 2\left( \frac{y}{2} \right) + 2 \]\[ x = (\frac{4y^2}{4}) - y + 2 \]\[ x = y^2 - y + 2 \]

Step 7: Final Iteration: Choose the Correct Relation

After eliminating \(t\), we found the equation relating \(x\) and \(y\) is \(x = y^2 - y + 2\). Comparing with the given options, (a) is the correct choice.

Key Concepts

Velocity Integration

When handling projectile motion problems, understanding velocity integration is crucial. This principle involves determining the position of a moving particle by integrating its velocity over time. The velocity functions provide the rates of change in each direction:

• For position along the x-axis, the velocity is given by \(v_x = 8t - 2\).
• For position along the y-axis, the velocity is \(v_y = 2\).

To find the position functions, we need to integrate each velocity function with respect to time. For the x-component, the integral of \(v_x\) results in a position function dependent on time. This helps us understand how far and in which direction the particle moves as time progresses. Similarly, integrating \(v_y\) gives us the position function for the y-component.
Understanding how these velocities describe motion is foundational. It allows us to predict future positions and understand past ones, which is essential in various fields like physics, engineering, and even space exploration.

Position Functions

Position functions are derived by integrating a velocity function over time. These functions describe the exact trajectory of a particle in the plane at any given moment:

• The position function for x, after integrating \(v_x = 8t - 2\), becomes \(x(t) = 4t^2 - 2t + C_1\).
• Similarly, for y, integrating \(v_y = 2\) results in \(y(t) = 2t + C_2\).

Constants \(C_1\) and \(C_2\) are determined using initial conditions, such as a known position at a specific time.
These initial conditions help solve for the constants, ensuring our position functions accurately represent the motion of the particle:

• For x: Given \(x = 14\) at \(t = 2\), we find that \(C_1 = 2\), making \(x(t) = 4t^2 - 2t + 2\).
• For y: Given \(y = 4\) at \(t = 2\), we find \(C_2 = 0\), resulting in \(y(t) = 2t\).

Accurate position functions are pivotal to mapping out the trajectory of moving bodies, allowing us to comprehend the dynamics in motion, such as energy changes and forces exerted.

Parameter Elimination

Parameter elimination is used to describe the motion path without using time as an intermediary variable. Once we have position functions, we can express one variable solely in terms of another by eliminating the time parameter \(t\).
From the function \(y(t) = 2t\), we can express \(t\) as \(t = \frac{y}{2}\). By substituting this into \(x(t) = 4t^2 - 2t + 2\), we can derive the relationship between \(x\) and \(y\) without involving \(t\). Performing this substitution gives us:

• \(x = 4\left(\frac{y}{2}\right)^2 - 2\left(\frac{y}{2}\right) + 2\)
• Simplifying, this becomes \(x = y^2 - y + 2\).

Parameter elimination is highly beneficial in mathematics and physics as it allows for an equation of a path directly relating different variables. This helps to describe paths and trajectories cleaned of unnecessary variables, simplifying analysis and solution finding in physics problems and more complex scenarios.