A body moves on a frictionless plane starting from rest. If $\mathrm{S}_{\mathrm{n}}$ is distance moved between $\mathrm{t}=\mathrm{n}-1$ and t $=n$ and $S_{n-1}$ is distance moved between $t=n-2$ and $\mathrm{t}=\mathrm{n}-1$, then the ratio $\frac{\mathrm{S}_{\mathrm{n}-1}}{\mathrm{~S}_{\mathrm{n}}}$ is $\left(1-\frac{2}{\mathrm{x}}\right)$ for n $=10$. The value of x is ______

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Kinematics

A body moves on a frictionless plane starting from rest. If

S_{n}

is distance moved between

t = n - 1

and t

= n

and

S_{n - 1}

is distance moved between

t = n - 2

and

t = n - 1

, then the ratio

\frac{S _{n - 1}}{S _{n}}

(1 - \frac{2}{x})

for n

= 10

. The value of x is ______

Answer: 19

Solution

For motion from rest, distance in the

n

-th second is

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

. Thus

\frac{S _{n - 1}}{S _{n}} = \frac{2 n - 3}{2 n - 1} = 1 - \frac{2}{2 n - 1} .

Comparing with

1 - \frac{2}{x}

gives

x = 2 n - 1

. For

n = 10

x = 19

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Integer answer

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