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A woodpecker鈥檚 brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker鈥檚 head comes to a stop from an initial velocity of 0.600 m/sin a distance of only 2.99 mm.

(a) Find the acceleration in m/s2 and in multiples of (g = 9.80 m/s2).

(b) Calculate the stopping time.

(c) The tendons cradling the brain stretch, making its stopping distance 4.50(greater than the head and, hence, less deceleration of the brain). What is the brain鈥檚 deceleration, expressed in multiples of g?

Short Answer

Expert verified

a) -90 m/s2,颅 -9.174 g.

b) 0.006 s.

c) -40 m/s2, -4.077 g.

Step by step solution

01

Given data

  • The initial velocity of the wood picker, U= 0.600 m/s.
  • Final velocity, V=0 m/s.
  • The distance of the wood picker, D= 2.00 mm = 0.002 m.
02

declaration and time period

a)deceleration of the woodpecker can be calculated using the equation as:

V2-U2=2ad

Here V is the final velocity U is the initial velocity a is the acceleration, and d is the distance traveled.

Substituting values in the above expression, we get,

(0)2-(0.6)2=2(a)(0.002)-0.36=2(0.002)aa=-0.362(0.002)a=-90m/s2

Hence the deceleration of the woodpecker is -90 m/s2.

b) The stopping time can be calculated as:

V = U + at

Here V is the final velocity, U is the initial velocity, a is the acceleration, and t is the time.

Substituting values in the above expression, we get,

0=0.6+(-90)tt=0.690t=0.006s

Hence the time of stopping is 0.006 s.

03

deceleration if the distance is 4.50 mm

c) The deceleration of the woodpecker can be calculated using the equation as:

V2-U2=2ad

Here V is the final velocity U is the initial velocity a is the acceleration, and d is the distance traveled.

Here d = 4.50 mm = 0.0045 m.

Substituting values in the above expression, we get,

(0)2-(0.6)2=2(a)(0.0045)-0.36=2(0.0045)aa=-0.362(0.0045)a=-40m/s2

Hence the Deceleration of the substance is -40 m/s2.

For the deceleration in multiple of g, we have to divide the deceleration by9.81.

The deceleration in multiple of g:

a=-40g9.81a=-4.077g

Hence the deceleration is -40 m/s2 and in multiple of g is -4.077 g.

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