A body cover a distance of 20 m in 7th s and 24 m in 9th s. How much distance will it cover 15th s?

A body cover a distance of 20 m in 7th second and 24 m in 9th second. How much distance will it cover in 15th second??

To solve this problem, we can use the concept of uniformly accelerated motion. We are given the distances covered by the body in the 7th and 9th seconds, and we need to find the distance covered in the 15th second.

Step-by-Step Solution:

1. *Distance covered in the nth second:*
\[
S_n = u + \frac{a}{2} \cdot (2n - 1)
\]
where:
- \( S_n \) = Distance covered in the nth second
- \( u \) = Initial velocity
- \( a \) = Acceleration
- \( n \) = nth second

2. *Equations for the 7th and 9th seconds:*

For the 7th second:
\[
S_7 = u + \frac{a}{2} \cdot (2 \times 7 - 1) = u + \frac{13a}{2}
\]
Given, \( S_7 = 20 \) m:
\[
20 = u + \frac{13a}{2} \quad \text{(Equation 1)}
\]

For the 9th second:
\[
S_9 = u + \frac{a}{2} \cdot (2 \times 9 - 1) = u + \frac{17a}{2}
\]
Given, \( S_9 = 24 \) m:
\[
24 = u + \frac{17a}{2} \quad \text{(Equation 2)}
\]

3. *Solve Equations 1 and 2:*

Subtract Equation 1 from Equation 2:
\[
(24 - 20) = \left(u + \frac{17a}{2}\right) - \left(u + \frac{13a}{2}\right)
\]
\[
4 = \frac{4a}{2} = 2a
\]
\[
a = 2 \text{ m/s}^2
\]

Now, substitute \( a = 2 \) in Equation 1:
\[
20 = u + \frac{13 \times 2}{2} = u + 13
\]
\[
u = 7 \text{ m/s}
\]

4. *Find the distance in the 15th second:*

\[
S_{15} = u + \frac{a}{2} \cdot (2 \times 15 - 1) = 7 + \frac{2 \times 29}{2} = 7 + 29 = 36 \text{ m}
\]

Final Answer:
The body will cover *36 meters* in the 15th second.

YouTube Description:

In this video, we solve a kinematics problem involving uniformly accelerated motion. We are given the distances covered by a body in the 7th and 9th seconds and use these to find the distance it covers in the 15th second. We apply the equation for distance covered in the nth second, solve for the acceleration and initial velocity, and finally determine the distance for the 15th second. Watch the video to see the step-by-step solution and understand the concepts of motion under constant acceleration.

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