Problem 22 - Master Resource Book in JEE Main Physics

Question

The displacement \((x)\) of a particle depends on time \((t)\) as $$ x=\alpha t^{2}-\beta t^{3} $$ (a) The particle will come to rest after time \(2 \alpha / 3 \beta\) (b) The particle will return to its starting point after time \(\alpha / \beta\) (c) No net force will act on the particle at \(t=\alpha / 3 \beta\) (d) The initial velocity of the particle was zero but its initial acceleration was not zero

Step-by-Step Solution

Answer

Options (a), (b), (c), and (d) are all correct.

Step 1: Calculate velocity

Velocity is the first derivative of displacement with respect to time. Given \( x = \alpha t^2 - \beta t^3 \), we differentiate to find velocity \( v(t) = \frac{dx}{dt} = 2\alpha t - 3\beta t^2 \).

Step 2: Find time when particle is at rest

The particle is at rest when its velocity is zero, i.e., \( v(t) = 0 \). This gives us the equation: \( 2\alpha t - 3\beta t^2 = 0 \). Factor out \( t \), giving \( t(2\alpha - 3\beta t) = 0 \). Thus, \( t = 0 \) or \( t = \frac{2\alpha}{3\beta} \).

Step 3: Determine time when the particle returns to starting point

The particle returns to its starting point when the displacement \( x = 0 \). Set \( \alpha t^2 - \beta t^3 = 0 \), which factors to \( t^2(\alpha - \beta t) = 0 \). Thus, \( t = 0 \) or \( t = \frac{\alpha}{\beta} \).

Step 4: Find time when net force is zero

Net force is zero when acceleration, which is the second derivative of displacement or the first derivative of velocity, is zero. Differentiate velocity: \( a(t) = \frac{dv}{dt} = 2\alpha - 6\beta t \). Set \( a(t) = 0 \) to find \( t = \frac{2\alpha}{6\beta} = \frac{\alpha}{3\beta} \).

Step 5: Determine initial conditions

Find initial velocity: substitute \( t = 0 \) in velocity expression: \( v(0) = 2\alpha \times 0 - 3\beta \times 0^2 = 0 \). Find initial acceleration: substitute \( t = 0 \) in acceleration expression: \( a(0) = 2\alpha - 6\beta \times 0 = 2\alpha eq 0 \).

Key Concepts

Velocity

Velocity is a measure of the speed and direction at which an object is moving. In kinematics, velocity is derived from the displacement of an object.,
For the problem at hand, displacement is given by the equation \( x = \alpha t^2 - \beta t^3 \).
To find the velocity, take the first derivative of the displacement with respect to time: \ \( v(t) = \frac{dx}{dt} = 2\alpha t - 3\beta t^2 \).
This equation tells us the velocity of the particle depends on both time and the constants \( \alpha \) and \( \beta \).
Understanding velocity is crucial as it helps determine when the particle is moving and when it comes to a rest.

Acceleration

Acceleration refers to how quickly the velocity of an object changes over time.
In other words, it's the derivative of velocity. For this situation, we calculate acceleration from our velocity function \( v(t) = 2\alpha t - 3\beta t^2 \) by taking the derivative: \ \( a(t) = \frac{dv}{dt} = 2\alpha - 6\beta t \).
This indicates that acceleration is constant initially and decreases linearly with time.

• When \( t = 0 \), \( a(0) = 2\alpha \), indicating initial acceleration is not zero.
• The acceleration becomes zero when \( t = \frac{\alpha}{3\beta} \), meaning no net force acts on the particle at that time.

Understanding acceleration helps us identify how forces affect the movement of the particle.

Displacement

Displacement is the term used in kinematics to describe an object's change in position.
It is not just about the distance traveled, but also includes direction.
In our example, the displacement of the particle is defined by the function \( x = \alpha t^2 - \beta t^3 \).
Here, the constants \( \alpha \) and \( \beta \) shape the curve of the particle's path over time. Evaluating displacement helps us solve when the particle returns to its starting point.
By setting \( x = 0 \), we find when the particle comes back: \( t = \frac{\alpha}{\beta} \).
This means the particle's journey brings it back to the start after a time duration of \( \frac{\alpha}{\beta} \).

Time of Rest

The time of rest is the point when the particle stops moving.
In other words, this is when the velocity is zero. From the velocity equation \( v(t) = 2\alpha t - 3\beta t^2 \), set \( v(t) = 0 \) to find these moments.
This leads to \( 2\alpha t - 3\beta t^2 = 0 \),which can be simplified to straight factors \( t(2\alpha - 3\beta t) = 0 \).
This results in two solutions for time: \( t = 0 \) or \( t = \frac{2\alpha}{3\beta} \).
This indicates the particle is at rest initially and once again at \( t = \frac{2\alpha}{3\beta} \). Knowing when a particle comes to a halt is pivotal in precisely predicting its motion.

Return to Starting Point

Understanding when a particle returns to its starting point involves setting displacement back to zero.
We look for the time when the initial and final positions are the same.
The movement is given by \( x = \alpha t^2 - \beta t^3 \).To determine when it returns to the start, set the displacement equal to zero: \( \alpha t^2 - \beta t^3 = 0 \), which factors to \( t^2(\alpha - \beta t) = 0 \).
Here, solutions are \( t = 0 \) and \( t = \frac{\alpha}{\beta} \).
The first solution, \( t = 0 \), suggests the starting point, and the second solution states the particle returns to its initial position after a time duration of \( \frac{\alpha}{\beta} \).
This insight is essential to understand full cyclical paths in particle motion.