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More # A body moving with uniform acceleration travels a distance ${s_n} = \left( {0.4n + 9.8} \right)\,m$ in ${n^{th}}$ second. Initial velocity of the body is (in $m{s^{ - 1}}$ )A. $0.4$B. $10$C. $3.5$D. $4$ Verified
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Hint: Here first we have to find the arithmetic progression for $n$ and $n - 1$ seconds. And then equate the result with the given progression to get the answer.

${s_n} = \left( {0.4n + 9.8} \right)\,m$
${s_n}$ distance travelled in ${n^{th}}$ and ${\left( {n - 1} \right)^{th}}$ seconds is:
Let ${u_1}$ be the initial velocity and $a$ be the acceleration
${s_n} = \left( {{u_1}n + \dfrac{1} {2}a{n^2}} \right) - \left( {{u_1}\left( {n - 1} \right) + \dfrac{1} {2}a{{\left( {n - 1} \right)}^2}} \right) \\ = {u_1} + \dfrac{1} {2}a\left( {{n^2} - {{\left( {n - 1} \right)}^2}} \right) \\ = {u_1} + \dfrac{a} {2}\left( { + 2n - 1} \right) \\ = {u_1} + an - \dfrac{a} {2} \\$
Comparing ${s_n}$ with the given equation we get,
$a = 0.4$

${u_1} - \dfrac{{0.4}} {2} = 9.8 \\ {u_1} = 10\,m{s^{ - 1}} \\$

Hence, initial velocity of the body is $10m{s^{ - 1}}$.

Therefore option B is correct.

Note:Here we have to be careful while calculating the initial velocity. We cannot write the initial velocity simply as $9.8$. We have to calculate from the equation of ${n^{th}}$ velocity.