College Physics by Openstax Chapter 2 Problem 32

A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm.

a) Find the acceleration in m/s2 and in multiples of g (g=9.80 m/s2).

b) Calculate the stopping time.

c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g?

Solution:

We are given the following values: v0=0.600 m/sv_0=0.600 \ \text{m/s}; vf=0.000 m/sv_f=0.000\:\text{m/s}; and Δx=0.002 m\Delta x=0.002\:\text{m}.

Part A

The acceleration is computed based on the formula,

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for acceleration aa in terms of the other variables, we have

a=(vf)2−(v0)22Δxa=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values,

a=(vf)2−(v0)22Δxa=(0.000 m/s)2−(0.600 m/s)22(0.002 m)a=−90.0 m/s2  (Answer)\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.002\:\text{m}\right)} \\ a & =-90.0\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

In terms of gg taking absolute values of the acceleration , we have

a=90.0 m/s2⋅(g9.80 m/s2)a=90.09.80ga=9.18g  (Answer)\begin{align*} a & = 90.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\ a & = \frac{90.0}{9.80}g \\ a & = 9.18g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We shall use the formula

Δx=vavet\Delta x=v_{ave}t

where vavev_{ave} is the average velocity computed as

vave=v0+vf2v_{ave}=\frac{v_0+v_f}{2}

Solving for time tt in terms of the other variables, we have

t=2Δxv0+vft=\frac{2\Delta x}{v_0+v_f}

Substituting the given values, we have

t=2Δxv0+vft=2(0.002 m)0.600 m/s+0.000 m/st=6.67×10−3 s  (Answer)\begin{align*} t & =\frac{2\Delta x}{v_0+v_f} \\ t & =\frac{2\left(0.002\:\text{m}\right)}{0.600\:\text{m/s}+0.000\:\text{m/s}} \\ t & =6.67\times 10^{-3}\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

Employing the same formula we used in Part A, we have

a=(vf)2−(v0)22Δxa=(0.000 m/s)2−(0.600 m/s)22(0.0045 m)a=−40.0 m/s2\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.0045\:\text{m}\right)} \\ a & =-40.0\:\text{m/s}^2 \end{align*}

In terms of gg, taking absolute values of the acceleration

a=40.0 m/s2⋅(g9.80 m/s2)a=40.09.80ga=4.08g  (Answer)\begin{align*} a & = 40.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\ a & =\frac{40.0}{9.80}g \\ a & = 4.08g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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