Kinematics

JEE Main level, numerical answer, medium

123 of 308

Question 123 of 308

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

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Answer: 19

Sn=12a(2n−1)=19a2Sn−1=12a(2n−3)=17a2Sn−1Sn=1719=1−2x⇒x=19\begin{aligned} & S_n=\frac{1}{2} a(2 n-1)=\frac{19 a}{2} \\ & S_{n-1}=\frac{1}{2} a(2 n-3)=\frac{17 a}{2} \\ & \frac{S_{n-1}}{S_n}=\frac{17}{19}=1-\frac{2}{x} \Rightarrow x=19\end{aligned}

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