JEE PYQ: Motion in a Straight Line - Question ID d4fe173e9217 (JEE Main 2011)

Core Formula & Concept:

This problem involves non-uniform deceleration—the deceleration is not constant but depends on the instantaneous speed \( v \). The given relation is: $\frac{dv}{dt} = -2.5 \sqrt{v}$ This is a first-order differential equation relating velocity and time. To find the time taken for the object to come to rest, we must solve this differential equation with the initial condition \( v(0) = 6.25 \, \text{m/s} \) and find the time \( t \) when \( v(t) = 0 \).

Key concepts used:

• Separation of variables in differential equations.
• Integration of power functions.
• Application of initial conditions to solve for constants.

--- Step-by-Step Derivation:

Step 1: Rewrite the differential equation

We are given: $\frac{dv}{dt} = -2.5 \sqrt{v}$ This can be rewritten as: $\frac{dv}{\sqrt{v}} = -2.5 \, dt$ This is a separable differential equation.

Step 2: Integrate both sides

Integrate the left side with respect to \( v \), and the right side with respect to \( t \): $\int \frac{dv}{\sqrt{v}} = \int -2.5 \, dt$ The left integral: $\int v^{-1/2} dv = 2 v^{1/2} + C_1$ The right integral: $\int -2.5 \, dt = -2.5 t + C_2$ So, combining: $2 \sqrt{v} = -2.5 t + C$ where \( C = C_2 - C_1 \) is the combined constant of integration.

Step 3: Apply the initial condition

At \( t = 0 \), \( v = 6.25 \, \text{m/s} \): $2 \sqrt{6.25} = -2.5 \cdot 0 + C \Rightarrow 2 \cdot 2.5 = C \Rightarrow C = 5$ So the equation becomes: $2 \sqrt{v} = -2.5 t + 5$

Step 4: Solve for \( t \) when \( v = 0 \)

When the object comes to rest, \( v = 0 \): $2 \sqrt{0} = -2.5 t + 5 \Rightarrow 0 = -2.5 t + 5$ $2.5 t = 5 \Rightarrow t = \frac{5}{2.5} = 2 \, \text{seconds}$

Conclusion:

The time taken by the object to come to rest is 2 seconds, which corresponds to option A. --- Common Traps & Exam Tip:

Trap 1: Misinterpreting the differential equation
Students often confuse \( \frac{dv}{dt} = -2.5 \sqrt{v} \) with constant deceleration. They might try to use \( v = u + at \), which is invalid here because acceleration is not constant. Always check if acceleration depends on velocity or time.

Trap 2: Incorrect integration
A common mistake is to integrate \( \frac{dv}{\sqrt{v}} \) incorrectly as \( \ln|\sqrt{v}| \) instead of \( 2\sqrt{v} \). Remember: $\int v^n dv = \frac{v^{n+1}}{n+1} + C \quad \text{for } n \neq -1$ Here, \( n = -1/2 \), so the integral is \( 2 v^{1/2} \).

Trap 3: Forgetting to apply initial conditions
Without applying \( v(0) = 6.25 \), the constant \( C \) remains unknown, and the solution cannot be completed. Always use initial conditions to solve for constants.

Exam Tip:
When acceleration depends on velocity, separation of variables is usually the way to go. Write \( \frac{dv}{f(v)} = g(t) dt \), then integrate both sides. This technique is frequently tested in JEE and should be practiced thoroughly.