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2015-03-04T16:34:53+05:30

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Distance travelled by a particle travelling with uniform acceleration(a) and initial velocity(u) in a straight line in the n^{th} second is given as

S_n=u+ \frac{a}{2} \times (2n-1)

given that S_8=55m\ and\ S_1_3=85m

S_8=u+ \frac{a}{2} \times (2 \times 8-1)\\\\ \Rightarrow55=u+ \frac{15a}{2}\\\\ \Rightarrow55=u+7.5a

Similarly\\S_1_3=u+ \frac{a}{2} \times (2 \times 13-1)\\ \\ \Rightarrow85=u+ \frac{25a}{2}\\\\ \Rightarrow85=u+12.5a

Subtracting second equation from the first, we get
30 = 5a
⇒ a = 30/5 = 6 m/s²

from the first equation,
55=u+7.5a\\ \\ \Rightarrow55 = u+7.5 \times 6\\ \\ \Rightarrow55 = u+45\\ \\  \Rightarrow u=55-45=10\ m/s

Acceleration is 6m/s² and the initial velocity is 10m/s.
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