Problem 1 The speed of propagation of the action potential (an electrical signal

Question

Problem 1 The speed of propagation of the action potential (an electrical signal) in a human nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance? Product 2 A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350m/s, how far will it travel before becoming airborne? (b) How long does this take?

Explanation / Answer

Problem 1:

We know tat

Speed = distance/time

Now given that

length of spinal cord to your feet = 1.1 m

Speed of nerve impulse = 18 m/sec, So

time taken by signal to travel this distance will be

time = distance/speed

time = 1.1 m/(18 m/s) = 1.1/18 sec

time = 0.0611 sec

Problem 2:

Given that

Initial Speed of swan = 0 m/sec

take off speed = 6.00 m/sec

acceleration of swan = 0.350 m/sec^2

Using 3rd kinematic equation

V^2 = U^2 + 2*a*S

S = (V^2 - U^2)/(2a)

S = (6^2 - 0^2)/(2*0.350)

S = 51.43 m = distance travelled by swan before becomin airborne

Part B.

time taken will be

Using 1st kinematic equation

V = U + a*t

t = [V - U]/a

t = [6 - 0]/0.350

t = 17.14 sec

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