Question

A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.470 m/s in a distance of only 2.30 mm. Find the acceleration in m/s² and in multiples of g (g = 9.80 m/s²)

accleration:

multiples of g:

calculate the stopping time:

The tendons cradling the brain stretch, making its stopping distance 5.06 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g?

Answer 1

Here ,

the cross product is

(2 i - 3j + 4 k ) X (A) = 4 i + 3j - k

Now, as the cross product vector is perpendicular to both the vectors

for a vector A to exist

(2 i - 3j + 4 k ) . ( 4 i + 3j - k ) must be zero

(2 i - 3j + 4 k ) . ( 4 i + 3j - k ) = 2 * 4 - 3 * 3 - 4

(2 i - 3j + 4 k ) . ( 4 i + 3j - k ) = -3

as the dot product is not zero , hence , these vectors are not perpendicular .

a) I do not beleive his claim

b) for the explanation given above

puneet • Thu, 8:10 AM

Here ,

intial speed , v = 0.470 m/s

distance , d = 2.3 mm = 0.0023 m

let the accleration is a

using third equation of motion

v^2 - u^2= 2* a * d

0 - 0.47^2 = 2 * 0.0023 * a

a = - 48 m/s^2 = - 4.9 * g

the acceleration of the head is -48 m/s^2 pr -4.9 * g

--------------------------------------------

stopping time = (v - u)/a

stopping time = (0 - 0.47)/(-48)

stopping time =0.00979 s

the stopping time is 0.00979 s

------------------------------------
d = 5.06 mm = 0.00506 m

let the accleration is a

using third equation of motion

v^2 - u^2= 2* a * d

0 - 0.47^2 = 2 * 0.00506 * a

a = - 21.8 m/s^2 = - 2.22 * g

the acceleration of the head is - 2.22 * g