運動學問題一物由靜止開始作等加速度...

Distance travelled in (n-1) seconds
= (a/2).(n-1)^2 [use equation: s = ut + (1/2)at^2 with u = 0 m/s]
Distance travelled in n second
= (a/2)n^2
Hence, distance travelled in the nth second
= (a/2)[n^2 - (n-1)^2]
= (a/2).(2n - 1)

Distance travelled in (n+1) seconds
= (a/2)(n+1)^2
Hence, distance travelled in the (n+1)th second
= (a/2).[(n+1)^2 - n^2]
= (a/2).(2n + 1)

Ratio of the two required distances
= [(a/2)(2n - 1)]/[(a/2)(2n + 1)]
= (2n - 1)/(2n + 1)

因為題目問第n秒，不是首n秒。