Answer
(a) $a = k~\frac{v^2}{r}$
(b) The acceleration increases by 21.0%
Work Step by Step
(a) Acceleration $a$ has the units $m/s^2$. In dimensional analysis, this would be expressed as $[L~T^{-2}]$.
Velocity $v$ has dimensions $[L~T^{-1}]$.
Radius $r$ has dimensions $[L]$.
We can write an equation for $a$ with a constant $k$:
$a = k~v^p~r^q$
We can use dimensional analysis to find $p$ and $q$:
$[L~T^{-2}] = [L^p~(T^{-1})^p] ~[L^q]$
$[L~T^{-2}] = [L^{p+q}~T^{-p}]$
We can see that $p = 2$.
Since $p+q=1$, then $q = -1$
We can write the equation for $a$:
$a = k~\frac{v^2}{r}$
(b) If the speed increases by 10.0%, then the new speed is $1.100v$. We can find the new expression for the acceleration:
$a' = k~\frac{(1.100v)^2}{r}$
$a' = 1.210~k~\frac{v^2}{r}$
The acceleration increases by 21.0%.
