During nth second of its motion a body covers a distance $S_n$ with uniform acceleration 'a' and initial velocity 'u'. Show that — $a = \dfrac{2S_n - 2u}{2n - 1}$

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Accepted Answer

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> [!TLDR]
> $S_n = s_n - s_{n-1} = u + 	frac{a}{2}(2n-1)$; solving for $a$ gives $a = \dfrac{2S_n - 2u}{2n-1}$.
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> The distance covered in $n$ seconds:
> $$s_n = un + 	frac12 a n^{2}.$$
> The distance covered in $(n-1)$ seconds:
> $$s_{n-1} = u(n-1) + 	frac12 a (n-1)^{2}.$$
> The distance covered in the $n$th second alone is the difference:
> $$S_n = s_n - s_{n-1} = u\big[n - (n-1)\big] + 	frac12 a\big[n^{2} - (n-1)^{2}\big].$$
> Now $n - (n-1) = 1$ and $n^{2} - (n-1)^{2} = (n + n - 1)(n - n + 1) = 2n - 1$, so …

Situation	Acceleration	Why it's (approximately) uniform
A car accelerating on a highway	~2–3 m/s²	Engine provides roughly constant force
A ball dropped from a height	9.8 m/s² downward	Gravity is nearly constant near Earth's surface
A train starting from a station	~0.5 m/s²	Controlled by the driver to be smooth

Q.During nth second of its motion a body covers a distance $S_n$ with uniform acceleration 'a' and initial velocity 'u'. Show that — $a = \dfrac{2S_n - 2u}{2n - 1}$

Concept understanding — Uniformly Accelerated Motion

Uniformly Accelerated Motion

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What It Looks Like in Real Life

The Key Insight

Q.During nth second of its motion a body covers a distance SnS_nSn​ with uniform acceleration 'a' and initial velocity 'u'. Show that — a=2Sn−2u2n−1a = \dfrac{2S_n - 2u}{2n - 1}a=2n−12Sn​−2u​

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Q.During nth second of its motion a body covers a distance $S_n$ with uniform acceleration 'a' and initial velocity 'u'. Show that — $a = \dfrac{2S_n - 2u}{2n - 1}$

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