Question

A body is moving in \(\mathrm{x}\) direction with constant acceleration \(\alpha\). Find the difference of the displacement covered by it in nth second and \((\mathrm{n}-1)\) th second. (A) \(\alpha\) (B) \((\alpha / 2)\) (C) \(3 \alpha\) (D) \((3 / 2) \alpha\)

Short answer

The difference of the displacement covered by the body in the nth second and (n-1)th second is \((\frac{3}{2})\alpha\).

Step-by-step solution

Understand the equations of motion

To solve this problem, we will use the equations of motion, which describe the relationship between the position, velocity, acceleration, and time for a body in motion. One of the equations we will use is derived from the initial velocity(u) and acceleration(a): \(v = u + at\)Where v is the final velocity and t is the time taken. Another equation of motion we'll use relates position(x), initial velocity(u), acceleration(a), and time(t): \(x = ut + \frac{1}{2}at^2\) We will use these equations to calculate the position at time t and then find the difference of the displacement covered by the body in the nth second and (n-1)th second.

Calculate the displacement at nth second and (n-1)th second

To calculate the displacement at nth second, we can use the equation for position: \(x_n = ut_n + \frac{1}{2}\alpha t_n^2\) Similarly, the displacement at (n-1)th second can be found using the equation for position: \(x_{n-1} = ut_{n-1} + \frac{1}{2}\alpha t_{n-1}^2\) It's important to notice that the displacement values found in these steps represent the total displacement from the start until the end of the nth second and (n-1)th second, respectively.

Finding the displacement covered in nth second and (n-1)th second

To find the displacement covered during the nth and (n-1)th seconds, we need to subtract the displacement that occurred during the previous seconds. Displacement covered during nth second: \(Δx_n = x_n - x_{n-1}\) Displacement covered during (n-1)th second: \(Δx_{n-1} = x_{n-1} - x_{n-2}\)

Calculate the difference of the displacement

Now, we can find the difference of the displacements covered during the nth second and (n-1)th second: \(Δx = Δx_n - Δx_{n-1}\) Substituting the values from Step 3: \(Δx = (x_n - x_{n-1}) - (x_{n-1} - x_{n-2})\) Using the position equations from Step 2: \(Δx = {(ut_n + \frac{1}{2}\alpha t_n^2) - (ut_{n-1} + \frac{1}{2}\alpha t_{n-1}^2)} - {(ut_{n-1} + \frac{1}{2}\alpha t_{n-1}^2) - (ut_{n-2} + \frac{1}{2}\alpha t_{n-2}^2)}\) Simplify the equation: \(Δx = {u(t_n - t_{n-1}) + \frac{1}{2}\alpha (t_n^2 - t_{n-1}^2)} - {u(t_{n-1} - t_{n-2}) + \frac{1}{2}\alpha (t_{n-1}^2 - t_{n-2}^2)}\) Since the time intervals are 1 second each: \(Δx = {u(1) + \frac{1}{2}\alpha (n^2 - (n-1)^2)} - {u(1) + \frac{1}{2}\alpha ((n-1)^2 - (n-2)^2)}\) Simplifying further: \(Δx = {\frac{1}{2}\alpha (n^2 - (n-1)^2 - (n-1)^2 + (n-2)^2)}\) Expanding and simplifying the terms: \(Δx = {\frac{1}{2}\alpha (2 - 3n + n^2 + n^2 - 4n + 4)}\)

Determine the answer from the equation

After simplification, we have: \(Δx = {(\frac{3}{2})\alpha}\) This corresponds to option (D). Therefore, the difference of the displacement covered by the body in the nth second and (n-1)th second is \((\frac{3}{2})\alpha\).

Key concepts

Constant Acceleration

In the world of physics, constant acceleration plays a significant role when analyzing motion. Constant acceleration means that the rate of change of velocity stays the same over time. Whether an object is speeding up or slowing down, the magnitude of its acceleration does not vary.
This concept is crucial as it simplifies how we can predict the motion of objects. When acceleration is constant, it allows us to use a set of equations known as the equations of motion to calculate different properties of motion.

Here's why constant acceleration is important:

• It allows for the prediction of future motion.
• It makes calculations straightforward since the acceleration value doesn’t change.
• It applies to a wide range of practical scenarios, from vehicles on a straight road to objects in free fall.

Understanding this concept helps lay the foundation for solving motion problems using kinematic equations.

Displacement Calculation

Displacement is a key concept in kinematics. It refers to the change in position of an object and is a vector quantity, meaning it has both magnitude and direction. In our problem, we're looking to calculate the difference in displacement for a moving body. This process involves leveraging the equation of motion that relates initial velocity, acceleration, and time.

When calculating displacement, especially with constant acceleration, you'll use the equation:

• For displacement at a specific second, you have: \(x = ut + \frac{1}{2}at^2\)
• Here, \(u\) represents initial velocity, \(a\) is acceleration, and \(t\) is time elapsed.

By plugging the right values into the equation, you can determine how far an object travels between specific time intervals in a certain direction. This is crucial for solving kinematic problems, especially when asked to find relative displacements, as in the exercise provided.

Kinematics

Kinematics is a branch of physics that focuses on the geometry of motion without taking into consideration the forces causing it. It describes how objects move using equations that relate time, displacement, velocity, and acceleration.

The foundational principles of kinematics involve several key equations known as the equations of motion:

• Velocity-time equation: \(v = u + at\)
• Displacement-time equation: \(x = ut + \frac{1}{2}at^2\)
• Velocity-displacement equation: \(v^2 = u^2 + 2ax\)

These equations assume constant acceleration and allow physicists and students to model and predict the motion of various objects under these conditions. By mastering these, you gain the insight needed to analyze any motion thoroughly, from a rolling ball to a speeding car.

Physics Problem Solving

Physics problem solving often requires a methodical approach. When faced with a problem like the one provided, involving motion and time, it's critical to identify which physics concepts and models apply.

Here’s a simple approach to solving such problems:

• Break down the problem into smaller parts.
• Identify known and unknown variables.
• Choose the right equations that link these variables.
• Solve step by step, checking units and results at each stage.

In the problem given, understanding the relationship between displacement, time, and constant acceleration is key. Tackling physics problems becomes less intimidating with practice and familiarity with the equations of motion. Treat these problems as puzzles. With time, dedication, and the right strategy, you will hone your problem-solving skills in physics.