JAMB - Physics (1992 - No. 4)

A body starts from rest and moves with a uniform acceleration of 6 ms \(^{-2}\). What distance does it cover in the third seconds
15m
18m
27m
30m

Explanation

\( u = 0 \, \text{m/s} \) (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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