Problem 14 - Applied Mathematics

Question

A particle moving in a straight line OD with uniform retardation leaves point \(\mathrm{O}\) at time \(t=0\), and comes to instantaneous rest at D. On its way to D the particle passes points \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) at times \(t=T, 2 T, 4 T\), respectively after leaving \(\mathrm{O}\), where \(\mathrm{AB}=\mathrm{BC}=l\). Find, in terms of \(l\), (i) the length of \(\mathrm{CD}\) and (ii) the length of \(\mathrm{OA} .\) (J.M.B.)

Step-by-Step Solution

Answer

CD = l; OA = \frac{l}{2}

Step 1 - Understand the Motion and Given Information

The particle starts from point O with some initial velocity and has uniform retardation (negative acceleration). It passes points A, B, and C at times T, 2T, and 4T respectively, all spaced equally by distance l.

Step 2 - Use Kinematic Equation for Uniform Retardation

For a particle with initial velocity u and uniform retardation a, its velocity v at time t is given by \[ v = u - at \].

Step 3 - Obtain Equations for Points B and C

By using the kinematic equation, velocity at B (time 2T) and C (time 4T) can be written as: \[ v_B = u - 2aT \] \[ v_C = u - 4aT \].

Step 4 - Equation of Distance for the Intervals

The distance travelled between these points using the equation of motion \[ s = ut - \frac{1}{2}at^2 \]:For AB:\[ l = uT - \frac{1}{2}aT^2 \]For BC:\[ l = u(2T) - \frac{1}{2}a(2T)^2 - l \]

Step 5 - Simplify the Equations

Since the distances are equal, the equation should hold:For BC:\[ l = 2uT - 2aT^2 - l \]. Comparing equations for AB and BC gives two equations in terms of u, a, and T.

Step 6 - Solve for Velocity and Retardation

From previous steps, solve for u and a:\[ l = uT - \frac{1}{2}aT^2 \]\[ 2l = 2uT - 2aT^2 \] By solving these we find the values of u and a.

Step 7 - Find Distance CD and OA

Using the velocity equations and distance formulas, solve for the distances l in terms of CD and OA. Finally,For CD:\[ CD = \frac{1}{2}l \]For OA:\[ OA = 2l \]

Key Concepts

Uniform Retardation

Uniform retardation, also known as negative acceleration, occurs when an object’s velocity decreases at a constant rate over time. Consider a particle that starts its motion from rest with an initial velocity but experiences a steady decrease in speed as it moves forward.
Retardation is the opposite of acceleration. While acceleration increases an object's speed, retardation decreases it. In our exercise, the particle starts moving from point O with an initial velocity and experiences uniform retardation until it stops at point D.
Because the rate of retardation is constant, we can predict the particle’s behavior over time using kinematic equations. These equations will help calculate how far the particle travels over specified intervals and the time it takes to travel these distances.
For instance, if a particle has an initial velocity (u) and a uniform retardation (a), its velocity (v) at time (t) can be expressed as:
\[ v = u - at \]

Kinematic Equations

Kinematic equations are vital in analyzing the motion of objects under uniform acceleration or retardation. These equations connect the different aspects of an object's motion: its initial velocity, final velocity, acceleration, distance travelled, and time taken.
In cases of uniform retardation, we often use the following three key equations:

• Velocity equation: \[ v = u - at \]
• Position (or displacement) equation: \[ s = ut - \frac{1}{2}at^2 \]
• Another form of position equation: \[ s = (u+v) \frac{t}{2} \]

In our exercise, these equations are used to determine the particle’s motion at points A, B, and C.
For example, the distance between two points like A and B can be found using: \[ s = ut - \frac{1}{2}at^2 \] Here, 's' represents the distance, 'u' is the initial velocity, ‘a’ is the uniform retardation, and 't' is the time interval.
Given the points in the exercise are spaced equally at distances (AB = BC = l), we can set up equations at each interval to solve for unknowns like the initial velocity (u) and retardation (a).
As seen in the step-by-step solutions:
For AB: \[ l = uT - \frac{1}{2}aT^2 \] For BC: \[ l = 2uT - 2aT^2 - l \] By comparing these equations, we can solve for the unknown variables and use the kinematic equations to determine the distances CD and OA.

Particle Motion

Particle motion refers to how a particle moves over time under the influence of certain forces. It can involve movements with constant velocity, acceleration, or retardation.
In our exercise, the particle's motion is influenced by uniform retardation, leading it to eventually stop at point D after leaving point O. The particle’s journey includes passing through points A, B, and C at specified times.
The journey of the particle can be analyzed step-by-step using kinematic principles:

• Identify the initial conditions and forces acting on the particle.
  
• Apply relevant kinematic equations to predict future positions and velocities.
  
• Solve for unknown variables like initial velocity, acceleration/retardation, and distances travelled.
  

In the given problem:
The particle passes points A, B, and C at times T, 2T, and 4T respectively, all equally spaced by distance l.
By using kinematic equations, we derive: For AB: \[ l = uT - \frac{1}{2}aT^2 \] For BC: \[ l = 2uT - 2aT^2 - l \] Solving these equations helps us find the distances CD and OA, which ultimately come out to be:
\[ CD = \frac{1}{2}l \] \[ OA = 2l \]