Problem Statement:
A car covers 2 m. in 2 seconds and 6 m. in the next 4 seconds. What distance will it cover in the 9th second?

Solution:

To find the distance covered in the 9th second, we need to analyze the given information and determine a pattern.

Given Information:
- The car covers 2 m. in 2 seconds.
- The car covers 6 m. in the next 4 seconds.

We can break down the given information as follows:

1. First 2 seconds:
- Distance covered: 2 m.

2. Next 4 seconds:
- Distance covered: 6 m.

From the given information, we can observe that the car is covering a greater distance in the next 4 seconds compared to the first 2 seconds. This indicates that the car is accelerating during this time.

Acceleration:
Acceleration is the rate at which an object changes its velocity over time. In this case, the car is accelerating because it covers a greater distance in the next 4 seconds compared to the first 2 seconds.

Constant Acceleration:
If an object has a constant acceleration, its distance covered can be determined using the following formula:

d = ut + 1/2at^2

Where:
- d is the distance covered
- u is the initial velocity
- a is the acceleration
- t is the time

Calculating Acceleration:
To determine the acceleration, we can use the information given in the problem statement.

1. First 2 seconds:
- Distance covered: 2 m.
- Time: 2 seconds.

Using the formula d = ut + 1/2at^2, we can substitute the given values:
2 = u(2) + 1/2a(2^2)

Simplifying the equation, we get:
2 = 2u + 2a

2. Next 4 seconds:
- Distance covered: 6 m.
- Time: 4 seconds.

Using the formula d = ut + 1/2at^2, we can substitute the given values:
6 = u(4) + 1/2a(4^2)

Simplifying the equation, we get:
6 = 4u + 8a

We now have a system of two equations with two variables (u and a). We can solve these equations simultaneously to find the values of u and a.

Solving the Equations:
Using the method of substitution or elimination, we can solve the system of equations:

Equation 1: 2 = 2u + 2a
Equation 2: 6 = 4u + 8a

By multiplying Equation 1 by 4 and Equation 2 by 2, we can eliminate the variable "u":

Equation 1 (multiplied by 4): 8 = 8u + 8a
Equation 2 (multiplied by 2): 12 = 8u + 16a

By subtracting Equation 1 from Equation 2, we can eliminate the variable "