Motion Analysis and Distance Calculation: An In-Depth Examination

Understanding motion and the factors that influence it is crucial in the realm of physics. This article delves into a specific problem involving distance and acceleration, providing a step-by-step guide to solving such problems and highlighting the importance of these calculations in real-world applications.

Problem Statement

A body covers 12 meters in the 2nd second and 20 meters in the 4th second. The question is: How much distance will it cover in the next 4 seconds after the 5th second?

Step-by-Step Solution

Step 1: Determine the Initial Conditions and Formulate Equations

To solve the problem, we need to determine the acceleration of the body and then calculate the distance it covers in the specified time intervals. We start by formulating the equations based on the distance covered in specific seconds.

We are given:

Distance covered in the 2nd second, ( S_2 12 ) meters Distance covered in the 4th second, ( S_4 20 ) meters

The distance covered in the ( n )th second can be expressed as:

[S_n u cdot frac{1}{2} a (2n - 1)]

Step 2: Set Up Equations for Given Data

For the 2nd second:

[S_2 u cdot frac{1}{2} a (2 cdot 2 - 1) u cdot frac{3}{2} a 12]

For the 4th second:

[S_4 u cdot frac{1}{2} a (2 cdot 4 - 1) u cdot frac{7}{2} a 20]

Step 3: Solve for Acceleration and Initial Velocity

From the first equation:

[u cdot frac{3}{2} a 12 implies u cdot 3a 24]

From the second equation:

[u cdot frac{7}{2} a 20 implies u cdot 7a 40]

Subtract the first equation from the second:

[(u cdot 7a - u cdot 3a) 40 - 24 implies 4ua 16 implies a 4 , text{m/s}^2]

Substitute ( a 4 ) m/s2 back into the first equation to find ( u ):

[u cdot 3 cdot 4 24 implies u 6 , text{m/s}]

Step 4: Calculate the Distance Covered From the 5th to the 9th Second

The distance covered in the first 5 seconds is given by:

[S_5 u cdot t frac{1}{2} a t^2 6 cdot 5 frac{1}{2} cdot 4 cdot 25 30 50 80 , text{meters}]

Now calculate the distance covered from the 6th second to the 9th second, which is 4 seconds:

For the 6th second:

[S_6 u cdot frac{1}{2} a (2 cdot 6 - 1) 6 cdot frac{1}{2} cdot 4 cdot 11 28 , text{meters}]

For the 7th second:

[S_7 u cdot frac{1}{2} a (2 cdot 7 - 1) 6 cdot frac{1}{2} cdot 4 cdot 13 32 , text{meters}]

For the 8th second:

[S_8 u cdot frac{1}{2} a (2 cdot 8 - 1) 6 cdot frac{1}{2} cdot 4 cdot 15 36 , text{meters}]

For the 9th second:

[S_9 u cdot frac{1}{2} a (2 cdot 9 - 1) 6 cdot frac{1}{2} cdot 4 cdot 17 40 , text{meters}]

Total distance covered from the 6th to the 9th second:

[S_6 S_7 S_8 S_9 28 32 36 40 136 , text{meters}]

Conclusion

The body will cover 136 meters in the 4 seconds after the 5th second. This example demonstrates the importance of understanding and applying the principles of motion analysis, particularly in scenarios involving acceleration and distance calculations.