A body starts from rest and moves with a uniform acceleration of 6 ms \(^{-2}\). What distance does it cover in the third seconds
\( u = 0 \, \text{m/s} \) (starts from rest)
\( a = 6 \, \text{m/s}^2 \)
Distance Covered in \( n \) Seconds
The distance covered in \( n \) seconds is given by:
\(d_n = u n + \frac{1}{2} a n^2\)
Since \( u = 0 \):
\(d_n = \frac{1}{2} a n^2 = \frac{1}{2} \cdot 6 \cdot n^2 = 3n^2\)
Calculate Distances
Distance covered in the first 2 seconds (\( d_2 \)):
\(d_2 = 3(2^2) = 3 \cdot 4 = 12 \, \text{m}\)
Distance covered in the first 3 seconds (\( d_3 \)):
\(d_3 = 3(3^2) = 3 \cdot 9 = 27 \, \text{m}\)
Distance Covered in the Third Second
The distance covered in the third second is the difference between the distances:
\(\text{Distance in the third second} = d_3 - d_2 = 27 \, \text{m} - 12 \, \text{m} = 15 \, \text{m}\)
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