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Q2.5-32PE

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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# 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 0.600 m/s in 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’s deceleration, expressed in multiples of g?

a) -90 m/s2,­ -9.174 g.

b) 0.006 s.

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

See the step by step solution

## 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.

## declaration and time period

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

${{\mathbit{V}}}^{{\mathbf{2}}}{\mathbf{-}}{{\mathbit{U}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{2}}{\mathbit{a}}{\mathbit{d}}$

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,

$\left(0{\right)}^{2}-\left(0.6{\right)}^{2}=2×\left(a\right)×\left(0.002\right)\phantom{\rule{0ex}{0ex}}-0.36=2×\left(0.002\right)×a\phantom{\rule{0ex}{0ex}}a=\frac{-0.36}{2×\left(0.002\right)}\phantom{\rule{0ex}{0ex}}a=-90{\text{m/s}}^{\text{2}}$

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,

$0=0.6+\left(-90\right)×t\phantom{\rule{0ex}{0ex}}t=\frac{0.6}{90}\phantom{\rule{0ex}{0ex}}t=0.006\text{s}$

Hence the time of stopping is 0.006 s.

## deceleration if the distance is 4.50 mm

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

${{\mathbit{V}}}^{{\mathbf{2}}}{\mathbf{-}}{{\mathbit{U}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{2}}{\mathbit{a}}{\mathbit{d}}$

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,

$\left(0{\right)}^{2}-\left(0.6{\right)}^{2}=2×\left(a\right)×\left(0.0045\right)\phantom{\rule{0ex}{0ex}}-0.36=2×\left(0.0045\right)×a\phantom{\rule{0ex}{0ex}}a=\frac{-0.36}{2\left(0.0045\right)}\phantom{\rule{0ex}{0ex}}a=-40{\text{m/s}}^{\text{2}}$

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

For the deceleration in multiple of g, we have to divide the deceleration by 9.81.

The deceleration in multiple of g:

$a=\frac{-40\text{g}}{9.81}\phantom{\rule{0ex}{0ex}}\text{a}=-4.077\text{g}$

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