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Q12PE

Expert-verifiedFound in: Page 82

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?**

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

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

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

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