Question

A car starts from rest and moves with constant acceleration.The ratio of distance covered by the car in nth second to that covered in n second is ​

Answer 1

Answer:

Given:

Car starts from rest and moves with constant acceleration.

To find:

Ratio of distance covered in n seconds to that of nth seconds.

Calculation:

Distance travelled in n seconds is :

$s1 = ut + \dfrac{1}{2} a {t}^{2}$

$= > s1 = 0 + \dfrac{1}{2} a {t}^{2}$

$= > s1 = \dfrac{1}{2} a {t}^{2}$

$= > s1 = \dfrac{1}{2} a {n}^{2}$

Distance travelled in the nth second is :

$s2 = u + \dfrac{1}{2} a(2n - 1)$

$= > s2 = 0+ \dfrac{1}{2} a(2n - 1)$

$= > s2 = \dfrac{1}{2} a(2n - 1)$

The above formula can be derived by subtraction of distance travelled in (n+1) seconds and distance travelled in (n) seconds.

Required ratio will be :

$\dfrac{s2}{s1} = \dfrac{ \dfrac{1}{2} a(2n -1 )}{ \dfrac{1}{2}a {n}^{2} }$

$= > s2 : s1 = (2n - 1):{n}^{2}$

So final answer :

$\boxed{ \red{ \sf{ \bold{ \huge{s2 : s1 = (2n - 1) : {n}^{2}}}}}}$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

Given :

• Initial velocity (u) = 0 m/s

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To Find :

• Ratio of Distance covered by car in nth second to the distance covered in n second.

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Solution :

We have formula for S nth second

$\large{\boxed{\sf{S_{n} \: = \: u \: + \: \dfrac{1}{2}(2n \: - \: 1)a}}} \\ \\ \implies {\sf{S_n \: = \: 0 \: + \: \dfrac{1}{2}(2n \: - \: 1)a}} \\ \\ \implies {\sf{S_n \: = \: \dfrac{1}{2}(2n \: - \: 1)a}}$

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Now, use formula for Distance traveled in n second

$\large{\boxed{\sf{S \: = \: ut \: + \: \dfrac{1}{2} at^2}}} \\ \\ \implies {\sf{S \: = \: 0 \: + \: \dfrac{1}{2}an^2}} \\ \\ \implies {\sf{S \: = \: \dfrac{1}{2} an^2}}$

$\rule{200}{2}$

Ratio is :

$\implies {\sf{\dfrac{S_n}{S} \: = \: \dfrac{\dfrac{1}{2}(2n \: - \: 1)a}{\dfrac{1}{2}an^2}}} \\ \\ \implies {\sf{\dfrac{S_n}{S} \: = \: \dfrac{(2n \: - \: 1)}{n^2}}} \\ \\ \implies {\sf{Ratio \: </p><p>\longrightarrow \: \boxed{\boxed{\sf{S_n \: : \: S \: = \: (2n \: - \: 1) \: : \: n^2}}}}}$