## College Physics (4th Edition)

(a) $a = k~\frac{v^2}{r}$ (b) The acceleration increases by 21.0%
(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%.