Solution:
Distance in \( n^{\text{th}} \) second is \( S_n = u + frac {a}{2}(2n-1)) \). Starting from rest \( u=0 , S_n =frac {a}{2}(2n-1).\) Given
\( S_n = 3 S_{n-1} \), we have 2n-1 = 3(2n-3). Solving gives (4n = 8), hence (n = 2).
At \(n^{\text{th}}\) second of the motion, the distance moved by the body is 3 times the distance moved in the previous second. The motion is uniformly accelerated & started from rest. The value of (n) is :
Leave a Reply