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A body covers a distance of 20 m in 7th sec and 24 m in the 9th sec. How much shall it cover in 15s ?

“A body covers a distance of 20 m in 7th s and 24 m in the 9th s. How much shall it cover in 15s ?”  – Raghav posted this question.

 

Answer: From the eqn for displacement in the nth second,

S_{n}=u + a\left ( n-\frac{1}{2} \right )

For the displacement in the 7th sec, we get,

20= u+a(7-\frac{1}{2})=u+a(\frac{13}{2}) …………………….. (1)

For the displacement in the 9th sec,

24= u+a(9-\frac{1}{2})=u+a(\frac{17}{2}) ……………………..(2)

(2)-(1) gives

24-20 = 2a

or a = 2 m/s2

Substituting for a in (1),

u = 7 m/s

Substituting, u= 7 m/s, a = 2 m/s2 and t = 15 sec in the eqn,
S = ut + \frac{1}{2} a t^{2}

we get S= 7 x 15 + 0.5 x 2x 15 x 15 = 105+225=330m

 

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