A body travels 102.5 m in nᵗʰ second and 115.0 m in (n+2)ᵗʰ second. The acceleration is :

Solution

Physics sir · 4 steps

Subtract the two nth-second distances and u cancels; the gap is two seconds, so it equals 2a.

1. Distance in nth second. For uniform acceleration, the distance covered in the nᵗʰ second is sₙ = u + a/2(2n - 1).
2. Subtract the two. Write the same formula for the (n+2)ᵗʰ second and subtract. The u terms cancel and so do the n terms, leaving only the gap of two seconds.

   sₙ₊₂ - sₙ = a/2[2(n+2) - 1 - (2n - 1)] = 2a

3. Put in numbers. 2a = 115.0 - 102.5 = 12.5 m, so a = 12.5/2 = 6.25 m s⁻².
4. So. The acceleration is 6.25 m s⁻², option (a).

The idea to keep. Distances in successive seconds go up by a each second. If the two seconds are k apart, their difference is ka, and you never need u or n.

Answer from the NTA final answer key. Solution by PrepWiser.