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Q2.5-32PE

Expert-verifiedFound in: Page 35

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/s ^{2} and in multiples of**

**(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/s ^{2}, -9.174 g.**

**b) 0.006 s.**

**c) -40 m/s ^{2}, -4.077 g.**

- 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.

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

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

**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,

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

Hence the deceleration of the woodpecker is -90 m/s^{2}.

**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,

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

Hence the time of stopping is 0.006 s.

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

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

**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,

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

Hence the Deceleration of the substance is -40 m/s^{2}.

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/s^{2} and in multiple of g is -4.077 g.

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