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Found in: Page 82

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fibre). If the nerve cell connecting the spinal cord to your feet is ${\mathbf{1}}{.}{\mathbf{1}}{\text{}}{\mathbf{m}}{\text{}}{\mathbf{long}}$, and the nerve impulse speed is ${\mathbf{18}}{\text{}}{\mathbf{m}}{/}{\mathbf{s}}$, how long does it take for the nerve signal to travel this distance?

$\mathbf{0}\mathbf{.}\mathbf{0611}\mathbf{\text{\hspace{0.17em}m/s}}$

See the step by step solution

## Step 1: Given Data

• Speed s= $\mathbf{18}\mathbf{\text{\hspace{0.17em}m/s}}$
• Distance d = $\mathbf{1}.\mathbf{1}\text{\hspace{0.17em}}\mathbf{m}$

## Step 2:  Time is taken to pass the impulse

Speed is a scalar quantity, whereas velocity is a vector quantity. Both need magnitude, but only velocity needs direction to express it.

Hence we can use the formula to solve the question:

${\mathbf{Time}}{\mathbf{=}}\frac{\mathbf{Distance}}{\mathbf{Speed}}$

Substituting the values in the above expression, we get:

$\begin{array}{c}\mathrm{T}=\frac{1.1\text{\hspace{0.17em}m}}{18\text{\hspace{0.17em}m/s}}\\ =0.0611\text{\hspace{0.17em}s}\end{array}$

Hence the time taken by the nerve cell is $\mathbf{0}\mathbf{.}\mathbf{0611}\mathbf{\text{\hspace{0.17em}m/s}}$ .