Chapter 2, Problem 86P

(a)

To determine

The average speed of the signal.

Answer to Problem 86P

The average speed of the signal is $1.0\text{mm/s}$,

Explanation of Solution

Write the equation for average speed of the signal.

$v_{\text{avg}}=\frac{\Delta x}{\Delta t}$ (I)

Here, $v_{\text{avg}}$ is the average speed, $\Delta x$ is the change in distance, and $\Delta t$ is the change in time.

Conclusion:

Substitute $100\text{nm}$ for $\Delta x$ and $\text{0}\text{.10}\text{ms}$ for $\Delta t$ in equation (I).

$\begin{array}{c}{v}_{\text{avg}}=\frac{100\text{nm}\left(\frac{{10}^{-9}}{1\text{n}}\right)}{\text{0}\text{.10}\text{ms}\left(\frac{{10}^{-3}}{1\text{m}}\right)}\\ =\frac{100\times {10}^{-9}\text{m}}{0.10\times {10}^{-3}\text{s}}\\ =1000\times {10}^{-6}\text{m/s}\end{array}$

Solve the above relation for velocity in terms of millimetre seconds.

$\begin{array}{c}{v}_{\text{avg}}=1.0\times {10}^{-3}\text{m/s}\left(\frac{1\text{mm}}{{10}^3\text{m}}\right)\\ =1.0\text{mm/s}\end{array}$

Therefore, the average speed of the signal is $1.0\text{mm/s}$,

(b)

To determine

The time taken by the signal to reach the brain.

Answer to Problem 86P

The time taken by the signal to reach the brain is $20\text{ms}$.

Explanation of Solution

Write the equation for time taken by the signal to the neuron.

$t_{\text{n}}=\frac{x}{v}$ (II)

Here, $t_{\text{n}}$ is the time taken by the signal to the neuron, $x$ is the distance pain signal travels into the neuron, and $v$ is the speed of the pain signal travels.

Write the equation for the total time travel across the synapses and neuron to reach the brain.

$t_{\text{tot}}=t_{\text{n}}+t_{\text{syn}}+t_{\text{n}}+t_{\text{syn}}$ (III)

Here, $t_{\text{tot}}$ is the total time travel to reach the brain, $t_{\text{n}}$ is the time taken by the signal to reach neuron, and $t_{\text{syn}}$ is the time taken by the synapses.

Rewrite the equation (III).

$\begin{array}{c}{t}_{\text{tot}}=2t_{\text{n}}+2t_{\text{syn}}\\ =2\left(t_{\text{n}}+t_{\text{syn}}\right)\end{array}$ (IV)

Conclusion:

Substitute $1.0\text{m}$ for $x$ and $\text{100}\text{m/s}$ for $v$ in equation (II).

$\begin{array}{c}{t}_{\text{n}}=\frac{1.0\text{m}}{\text{100}\text{m/s}}\\ =0.01\text{s}\\ =10\times {10}^{-3}\text{s}\left(\frac{1\text{ms}}{{10}^3\text{s}}\right)\\ =10\text{ms}\end{array}$

Substitute $10\text{ms}$ for $t_{\text{n}}$ and $0.10\text{ms}$ for $t_{\text{syn}}$ in equation (IV).

$\begin{array}{c}{t}_{\text{tot}}=2\left(10\text{ms}+0.10\text{ms}\right)\\ =20.2\text{ms}\\ =20\text{ms}\end{array}$

Therefore, the time taken by the signal to reach the brain is $20\text{ms}$.

(c)

To determine

The average speed of propagation of signal.

Answer to Problem 86P

The average speed of propagation of signal is $99\text{m/s}$.

Explanation of Solution

Write the equation for average speed of the signal.

${v^{\prime }}_{\text{avg}}=\frac{\Delta x}{\Delta t}$ (V)

Here, $v_{\text{avg}}$ is the average speed of the propagation signal, $\Delta x$ is the change in distance, and $\Delta t$ is the change in time.

Conclusion:

Substitute $2.0\text{m}+2\left(100\text{nm}\right)$ for $\Delta x$ and $20.2\text{ms}$ for $\Delta t$ in equation (V).

$\begin{array}{c}{v^{\prime }}_{\text{avg}}=\frac{2.0\text{m}+2\left(100\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{20.2\text{ms}\left(\frac{{10}^{-3}\text{s}}{1\text{ms}}\right)}\\ =\frac{2.0\text{m}+200\times {10}^{-9}\text{m}}{20.2\times {10}^{-3}\text{s}}\\ =\frac{2.0\text{m}+0.0000002\text{m}}{20.2\times {10}^{-3}\text{s}}\\ =0.099\times {10}^3\text{m/s}\end{array}$

Solve the above relation for velocity.

${v^{\prime }}_{\text{avg}}=99\text{m/s}$

Therefore, the average speed of propagation of signal is $99\text{m/s}$.