AP EAMCET · PHYSICS · Motion In One Dimension
A body starts from rest with uniform acceleration and its velocity at a time of ' n ' seconds is ' \(v\) '. The total displacement of the body in the \(n^{\text {th }}\) and \((n-1)^{\text {th }}\) seconds of its motion is
- A \(\frac{\mathrm{v}(\mathrm{n}+1)}{\mathrm{n}}\)
- B \(\frac{2 \mathrm{v}(\mathrm{n}+1)}{\mathrm{n}}\)
- C \(\frac{2 \mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
- D \(\frac{\mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
Step-by-step Solution
Detailed explanation
\(v = u + an\) \(v = 0 + an \implies a = \frac{v}{n}\) \(S_t = u + \frac{a}{2}(2t - 1)\) \(S_n = 0 + \frac{a}{2}(2n - 1)\) \(S_{n-1} = 0 + \frac{a}{2}(2(n-1) - 1) = \frac{a}{2}(2n - 3)\)…
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