# College Physics 2.32 – Acceleration of a woodpecker’s head

## Solution:

### Part A

The acceleration is computed based from the formula, $\displaystyle \text{v}^2=\text{v}_0^2+2\text{a}\Delta \text{x}$. If we solve for acceleration, we have

$\displaystyle \text{a}=\frac{\text{v}^2-\text{v}_0^2}{2\Delta \text{x}}$

$\displaystyle \text{a}=\frac{\left(0\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(2.00\times 10^{-3}\right)\text{m}}$

$\text{a}=-90.0\:\text{m/s}^2$

In terms of g.

$\displaystyle \frac{\left|a\right|}{\left|g\right|}=\frac{90.0\:\text{m/s}^2}{9.80\:\text{m/s}^2}$

$\displaystyle \frac{\left|a\right|}{\left|g\right|}=9.18$

Therefore,

$a=9.18\text{g}$

### Part B

From the formula $\Delta \text{x}=\frac{1}{2}\left(\text{v}_0+\text{v}\right)\text{t}$, we have

$\displaystyle \text{t}=\frac{2\Delta \text{x}}{\text{v}_0+\text{v}}$

$\displaystyle \text{t}=\frac{2\left(2.00\times 10^{-3}\:\text{m}\right)}{0.600\:\text{m/s}+0\:\text{m/s}}$

$\text{t}=6.67\times 10^{-3}\:\text{s}$

### Part C

$\displaystyle a=\frac{v^2-v_0^2}{2\Delta x}$

$\displaystyle a=\frac{\left(0\:m/s\right)^2-\left(0.600\:m/s\right)^2}{2\left(4.50\times 10^{-3}\:m\right)}$

$\displaystyle a=-40.0\:m/s^2$

In terms of g.

$\displaystyle \frac{\left|a\right|}{\left|g\right|}=\frac{40.0\:m/s^2}{9.80\:m/s^2}$

$\displaystyle \frac{\left|a\right|}{\left|g\right|}=4.08$

Therefore,

$\displaystyle a=4.08g$