Understanding Particle Motion with Constant Acceleration

This question asks us to analyze the motion of a particle that starts from rest and experiences constant acceleration for a total duration of 20 seconds. We need to find the relationship between the distance covered in the first 10 seconds and the distance covered in the subsequent 10 seconds.

Key Concepts for Constant Acceleration Problems

When a particle moves with constant acceleration, we can use the kinematic equations. The relevant equation here, since the particle starts from rest and we're dealing with distance, initial velocity, time, and acceleration, is:

\(s = ut + \frac{1}{2}at^2\)

Where:

• \(s\) is the distance traveled.
• \(u\) is the initial velocity.
• \(a\) is the constant acceleration.
• \(t\) is the time taken.

Analyzing the Particle's Motion

The problem provides the following information:

• Initial velocity, \(u = 0\) (starts from rest).
• Acceleration is constant, let's denote it by \(a\).
• Total time of motion considered is 20 seconds.
• \(X_1\) is the distance traveled in the first 10 seconds.
• \(X_2\) is the distance traveled in the next 10 seconds (from \(t=10\) s to \(t=20\) s).

Calculating the Distance in the First 10 Seconds ($X_1$)

We apply the kinematic equation for the first 10 seconds:

\(t = 10\) s

\(u = 0\)

\(a = a\)

So, the distance \(X_1\) is:

\(X_1 = (0)(10) + \frac{1}{2}a(10)^2\)

\(X_1 = 0 + \frac{1}{2}a(100)\)

\(X_1 = 50a\)

This gives us an expression for \(X_1\) in terms of the constant acceleration \(a\).

Calculating the Total Distance in 20 Seconds

Next, let's find the total distance traveled in the entire 20-second period. We apply the kinematic equation for the first 20 seconds:

\(t = 20\) s

\(u = 0\)

\(a = a\)

Let the total distance in 20 seconds be \(S_{20}\). Then:

\(S_{20} = (0)(20) + \frac{1}{2}a(20)^2\)

\(S_{20} = 0 + \frac{1}{2}a(400)\)

\(S_{20} = 200a\)

This is the total distance covered from \(t=0\) to \(t=20\) s.

Calculating the Distance in the Remaining 10 Seconds ($X_2$)

The distance \(X_2\) is the distance covered from the end of the first 10 seconds (\(t=10\) s) to the end of the 20 seconds (\(t=20\) s). This distance is the total distance covered in 20 seconds minus the distance covered in the first 10 seconds.

\(X_2 = S_{20} - X_1\)

Substitute the values we found for \(S_{20}\) and \(X_1\):

\(X_2 = 200a - 50a\)

\(X_2 = 150a\)

So, the distance \(X_2\) is \(150a\).

Finding the Relationship Between $X_1$ and $X_2$

We have the following expressions:

• \(X_1 = 50a\)
• \(X_2 = 150a\)

We can see that \(150a\) is three times \(50a\). Therefore, we can write the relationship as:

\(X_2 = 3 \times (50a)\)

\(X_2 = 3 X_1\)

This shows that the distance traveled in the second 10-second interval is three times the distance traveled in the first 10-second interval for a particle starting from rest under constant acceleration.

Summary of Distances

Let's summarize the distances calculated:

From the table, it is clear that \(X_2 = 3X_1\).

Revision Table: Constant Acceleration Motion

Additional Information: Galileo's Law of Odd Numbers

This problem illustrates a specific case of Galileo's Law of Odd Numbers. When a particle starts from rest and moves with constant acceleration, the distances covered in equal successive intervals of time are in the ratio of odd numbers: 1:3:5:7, and so on.

In this question, the first interval is 10 s, and the second interval is also 10 s (from 10 s to 20 s). Since the particle starts from rest and has constant acceleration:

• Distance in the 1st interval (\(X_1\)): Corresponds to the '1' in the ratio.
• Distance in the 2nd interval (\(X_2\)): Corresponds to the '3' in the ratio.

Therefore, the ratio \(X_1 : X_2\) is 1 : 3, which means \(X_2 = 3X_1\).

Let's verify this with our results:

• \(X_1 = 50a\)
• \(X_2 = 150a\)

The ratio \(X_1 / X_2 = (50a) / (150a) = 50/150 = 1/3\). So, \(X_2 = 3X_1\).

This law is a useful shortcut for problems involving motion from rest with constant acceleration over equal time intervals.