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Q. 1.2

The acceleration a of a particle is related to its velocity v, its position coordinate x, and time t by the equation

a = Ax³t + Bvt²                      (a)

where A and B are constants. The dimension of the acceleration is length per unit time squared; that is, [a] = [L/T²]. The dimensions of the other variables are [v] = [L/T], [x] = [L], and [t] = [T]. Derive the dimensions of A and B if Eq. (a) is to be dimensionally homogeneous.

Verified Solution

For Eq. (a) to be dimensionally homogeneous, the dimension of each term on the right-hand side of the equation must be [L/T]², the same as the dimension for a. Therefore, the dimension of the first term on the right-hand side of Eq. (a) becomes

$[Ax^{3}t] = [A][x^{3}][t]=[A][L^{3}][T] =\left[\frac{L}{T^{2}} \right]$                      (b)

Solving Eq. (b) for the dimension of A, we find

$[A] =\frac{1}{[L^{3}][T]}\left[\frac{L}{T^{2}} \right]=\frac{1}{\left[L^{2}T^{3}\right] }$

Performing a similar dimensional analysis on the second term on the righthand side of Eq. (a) gives

$[Bvt^{2}]=[B][v][t^{2}]=[B]\left[\frac{L}{T} \right]\left[T^{2}\right]=\left[\frac{L}{T^{2}} \right]$                      (c)

Solving Eq. (c) for the dimension of B, we find

$[B]=\left[\frac{L}{T^{2}}\right]\left[\frac{T}{L}\right]\left[\frac{1}{T^{2}} \right]=\left[\frac{1}{T^{3}} \right]$