Q12PE - College Physics (Urone)

Question

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  $\mathbf{1}.\mathbf{1}\mathbf{m}\mathbf{long}$, and the nerve impulse speed is $\mathbf{18}\mathbf{m}/\mathbf{s}$, how long does it take for the nerve signal to travel this distance?

Step-by-Step Solution

Answer

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

Step 1: Given Data

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

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