QJEEM24Phcc2d: JEE Main 2024 Physics - Class 11 JEEM24Phcc2d Solution: A body moves on a frictionless plane starting from rest. If is…

Given that a body is moving on a frictionless plane and starts from rest, the motion can be assumed to be uniformly accelerated motion. The formula for the distance covered in uniformly accelerated motion from rest is given by $s = ut + \frac{1}{2}at^2$ , where:

• $s$ is the distance covered,


• $u$ is the initial velocity (which is 0 since the body starts from rest),


• $a$ is the acceleration, and


• $t$ is the time.

Since the body starts from rest ( $u=0$ ), the formula simplifies to $s = \frac{1}{2}at^2$ .

The distance moved between $t = n - 1$ and $t = n$ , denoted as $S_n$ , can be found by calculating the distance covered by the end of time $n$ and subtracting the distance covered by the end of time $n-1$ . Let's denote the total distance covered by time $n$ as $S(n)$ , which according to the formula is $S(n) = \frac{1}{2}a n^2$ . Thus, $S_n = S(n) - S(n-1)$ .

Therefore, we have:

$S_n = \frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2$

Simplifying this, we get:

$S_n = \frac{1}{2}a(n^2 - (n^2 - 2n + 1))$

This simplifies to:

$S_n = \frac{1}{2}a(2n - 1)$

Similarly,

$S_{n-1} = \frac{1}{2}a((n-1)^2 - (n-2)^2)$

Simplifying:

$S_{n-1} = \frac{1}{2}a((n-1)^2 - (n^2 - 4n + 4))$

Which further simplifies to:

$S_{n-1} = \frac{1}{2}a(2n - 3)$

Hence, the ratio $\frac{S_{n-1}}{S_n}$ can be calculated:

$\frac{S_{n-1}}{S_n} = \frac{\frac{1}{2}a(2n - 3)}{\frac{1}{2}a(2n - 1)} = \frac{2n - 3}{2n - 1}$

For $n = 10$ ,

$\frac{S_{n-1}}{S_n} = \frac{2(10) - 3}{2(10) - 1} = \frac{20 - 3}{20 - 1} = \frac{17}{19}$

Given that the ratio is represented as $\left(1 - \frac{2}{x}\right)$ , we have:

$\frac{17}{19} = 1 - \frac{2}{x}$

Solving this equation for $x$ gives:

$1 - \frac{17}{19} = \frac{2}{x}$

$\frac{2}{19} = \frac{2}{x}$

Thus:

$x = 19$