Consider a grey squirrel falling out of a tree to the ground.

(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.

(b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

Solution:

Part A

We are given the following: v_0=0 \ \text{m/s}; a=-9.80 \ \text{m/s}^2; and \Delta x=-3.0\ \text{m}.

Note that the acceleration is due to gravity and its value is constant at a=-9.80 \ \text{m/s}^2. Also, the distance x, is negative because of the direction of motion. For free-fall, downward motion is considered negative. To solve for the velocity just before it hits the ground, we will solve v_fin the formula

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

Substituting the given values:

\begin{align*}
\left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\
v_f & = \sqrt{\left(0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-3.0\:\text{m}\right)} \\
v_f & = 7.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are given the following: v_f=0 \ \text{m/s}; v_0=7.7 \ \text{m/s}; and \Delta x=0.02 \ \text{m}

To solve for the acceleration we shall use the same formula as that in Part A. Solving for the acceleration a:

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & =\frac{\left(0\:\text{m/s}\right)^2-\left(7.7\:\text{m/s}\right)^2}{2\left(0.02\:\text{m}\right)} \\
a & =-1.5\times 10^3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

This is approximately 3 times the deceleration of the airman from the previous problem, who was falling from thousands of meters high.

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m.

a) What is his deceleration?

b) How long does the collision last?

Solution:

We are given the following: v_0=7.50\:\text{m/s}; v_f=0.00\:\text{m/s}; and\Delta x=0.350\:\text{m}.

Part A

We are going to use the formula

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

Solving for the acceleration a in terms of the other variables:

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & =\frac{\left(0\:\text{m/s}\right)^2-\left(7.50\:\text{m/s}\right)^2}{2\left(0.350\:\text{m}\right)} \\
a & =-80.4\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are going to use the formula

\Delta x=v_{ave}t

Since v_{ave}=\frac{v_f+v_0}{2}, we can write the formula as

\Delta x=\frac{v_f+v_0}{2}\cdot t

Solving for time t in terms of the other variables:

\:t=\frac{2\Delta x}{v_f+v_0}

Substituting the given values:

\begin{align*}
t&=\frac{2\Delta x}{v_f+v_0} \\
t&=\frac{2\left(0.350\:\text{m}\right)}{0\:\text{m/s}+7.50\:\text{m/s}} \\
t& =9.33\times 10^{-2}\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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: v_0=0.600 \ \text{m/s}; v_f=0.000\:\text{m/s}; and \Delta x=0.002\:\text{m}.

Part A

The acceleration is computed based on the formula,

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

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

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values,

\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 g taking absolute values of the acceleration , we have

\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

\Delta x=v_{ave}t

where v_{ave} is the average velocity computed as

v_{ave}=\frac{v_0+v_f}{2}

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

t=\frac{2\Delta x}{v_0+v_f}

Substituting the given values, we have

\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

\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 g, taking absolute values of the acceleration

\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*}