Question

The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position as x = kamtn, where k is a dimensionless constant. Show by dimensional analysis that this expression is satisfied if m = 1 and n = 2. (Submit a file with a maximum size of 1 MB.)

Answer 1

Answer:

Explanation:

$x = k\times a^m\times t^n$

k is constant , it is dimension is zero. Using dimensional unit , we cal write the relation as follows

L = $(LT^{-2})^m(T)^n$

= $L^mT^{-2m+n}$

Equating power of like items

m=1

-2m+n = 0

n = 2