Problem 119 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity \(\vec{u}\) and the other from rest with uniform acceleration \(\vec{f}\). Let \(\alpha\) be the angle between their directions of motion. The relative velocity of the second particle w.r.t. the first is least after a time (a) \(\frac{u \cos \alpha}{f}\) (b) \(\frac{u \sin \alpha}{f}\) (c) \(\frac{f \cos \alpha}{u}\) (d) \(u \sin \alpha\)

Step-by-Step Solution

Answer

The correct option is (a) \(\frac{u \cos \alpha}{f}\).

Step 1: Understand the Problem

We need to find the time at which the relative velocity of the second particle, which is accelerating, with respect to the first particle, which is moving with a constant velocity, is minimized.

Step 2: Relative Velocity Formula

The velocity of the first particle is represented as \( \vec{u} \). The velocity of the second particle, starting from rest with uniform acceleration \( \vec{f} \), is given by \( \vec{f}t \) at time \( t \). The relative velocity \( \vec{v}_{rel} \) of the second particle with respect to the first is \( \vec{f}t - \vec{u} \).

Step 3: Express Relative Velocity in Terms of Components

Considering the angle \( \alpha \) between \( \vec{u} \) and \( \vec{f} \), we can express \( \vec{u} \) as \( (u \cos \alpha, u \sin \alpha) \) and \( \vec{f} \) as \( (f \cos \alpha, f \sin \alpha) \). The relative velocity vector is \( (f t \cos \alpha - u \cos \alpha, f t \sin \alpha - u \sin \alpha) \).

Step 4: Find the Magnitude of Relative Velocity

The magnitude of the relative velocity \( v_{rel} \) is given by: \[ v_{rel} = \sqrt{(f t \cos \alpha - u \cos \alpha)^2 + (f t \sin \alpha - u \sin \alpha)^2} \] Expanding the squares can help in simplification.

Step 5: Optimize the Magnitude

We aim to minimize the magnitude. Taking the derivative with respect to \( t \) and setting it to zero will give the critical points:\[ \frac{d}{dt} \left( \sqrt{(f t \cos \alpha - u \cos \alpha)^2 + (f t \sin \alpha - u \sin \alpha)^2} \right) = 0 \]

Step 6: Solve for Minimum Time

Solving the derivative condition for \( t \), simplifying, we get the minimum time \( t \) as the time when the derivative equals zero. This is given by:\[ t = \frac{u \cos \alpha}{f} \] which corresponds to option (a).

Key Concepts

Uniform Velocity

Uniform velocity means that an object moves at a constant speed in a straight line.
This implies that both the speed and the direction of the object remain unchanged.
This simplicity makes it an excellent starting point when learning motion physics. Take a car driving straight down the highway at 60 miles per hour. Its rate of travel does not change, nor does the direction it is moving in.

Characteristics of uniform velocity include:

• Constant speed: The magnitude of the velocity remains the same over time.
• Straight-line motion: The direction of motion does not change.
• No acceleration: Since there is no speed or direction change, acceleration is zero.

Understanding uniform velocity helps students predict how far an object will travel in a given amount of time, using the simple formula: \[ \text{Distance} = \text{Velocity} \times \text{Time}\] This basic concept is crucial when comparing different types of motion, as seen in problems involving relative velocity, where a constant velocity component, such as \( \vec{u} \) in our exercise, plays a role in determining how the motion of another object compares over time.

Uniform Acceleration

Uniform acceleration occurs when the rate of change of velocity of an object is constant.
This means that the velocity changes by the same amount in each equal time period.
Imagine a car that speeds up by 10 miles per hour every second. Here, the car is experiencing uniform acceleration.

Some properties of uniform acceleration include:

• Constant rate of velocity change: The rate at which speed increases (or decreases) remains the same.
• Predictable motion: Using kinematic equations, future velocity and position can be easily calculated.
• Linear velocity-time graph: The slope of this graph indicates the acceleration.

For objects starting from rest, the formula for determining the velocity at any time \( t \) is:\[ \text{Velocity} = \text{Initial Velocity} + \text{Acceleration} \times \text{Time}\]This becomes \( \vec{f}t \) in the problem, where \( \vec{f} \) represents the constant acceleration vector.
This is a fundamental idea when calculating relative velocity between two objects, particularly when one is accelerating.

Angle Between Vectors

The angle between vectors affects their relative motion and is crucial in solving problems involving relative velocity.
The angle, denoted as \( \alpha \), is the measure between the direction vectors \( \vec{u} \) and \( \vec{f} \).
Think of it as the separation in direction between two arrows originating from the same point.

Key aspects of the angle between vectors:

• Dot product: The cosine of the angle between two vectors is involved in the dot product formula.
• Magnitude influences: The angle affects how components of vectors add or subtract.
• Relative positioning: By knowing the angle, one can determine how projections of one vector fall on another.

The dot product formula, which helps find the angle, is:\[ \vec{u} \cdot \vec{f} = \lvert \vec{u} \rvert \lvert \vec{f} \rvert \cos \alpha\]In our problem, understanding this angle is imperative for breaking vectors into components that aid in calculating relative velocity. When vectors are expressed in terms of \( cos \alpha \) and \( sin \alpha \), it facilitates evaluating how motions align or differ over time, impacting minimum relative velocity calculations.