a particle with velocity u is moving with uniform accelration a. Distance moved by particle in the nth second is given by s= ut+a)2t-1. What are the dimensions of s?

To determine the dimensions of a physical quantity, we analyze the dimensions of the various parameters involved. In this case, we will analyze the dimensions of the equation s = ut + (a/2)t^2.

Let's break down the dimensions of each term in the equation:

1. ut: This term represents the initial velocity u multiplied by time t. The dimensions of velocity are length/time, denoted as [LT^-1]. Therefore, the dimensions of ut would be [LT^-1] x [T] = [L].

2. (a/2)t^2: This term represents the acceleration a divided by 2, multiplied by time t squared. The dimensions of acceleration are length/time^2, denoted as [LT^-2]. Therefore, the dimensions of (a/2)t^2 would be [LT^-2] x [T]^2 = [L].

Combining the two terms, we have s = ut + (a/2)t^2 = [L] + [L] = [L], meaning that the dimensions of s are simply length, denoted as [L].

Therefore, the dimensions of the distance moved by the particle, s, are length ([L]).