Chapter 1 - Introduction to Physics - Problems and Conceptual Exercises - Page 17: 50

$p=1$ and $q=-1$

Dimension of t: [T] Dimension of v: $\frac{[L]}{[T]}$ Dimension of a: $\frac{[L]}{[T]^{2}}$ $a=v^{p}t^{q}$ $\frac{[L]}{[T]^{2}}=(\frac{[L]}{[T]})^{p}([T])^{q}$ $[L]^{1}\times[T]^{-2}=[L]^{p}\times[T]^{q-p}$ $p=1$ $q-p=-2$ $q-1=-2$ $q=-1$