# College Physics by Openstax Problem 2.32

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

SOLUTION:

Part a

$a=\frac{v^2-v_0^2}{2\Delta x}$

$a=\frac{\left(0\:m/s\right)^2-\left(0.600\:m/s\right)^2}{2\left(2.00\times 10^{-3}\:m\right)}$

$a=-90.0\:m/s^2$

In terms of g.

$\frac{\left|a\right|}{\left|g\right|}=\frac{90.0\:m/s^2}{9.80\:m/s^2}$

$\frac{\left|a\right|}{\left|g\right|}=9.18$

Therefore,

$a=9.18g$

Part b

From the formula $\Delta x=\frac{1}{2}\left(v_0-v\right)t$, we have

$t=\frac{2\Delta x}{v_0+v}$

$t=\frac{2\left(2.00\times 10^{-3}\:m\right)}{\left(0.600\:m/s\right)+\left(0\:m/s\right)}$

$t=6.67\times 10^{-3}\:s$

Part c

$a=\frac{v^2-v_0^2}{2\Delta x}$

$a=\frac{\left(0\:m/s\right)^2-\left(0.600\:m/s\right)^2}{2\left(4.50\times 10^{-3}\:m\right)}$

$a=-40.0\:m/s^2$

In terms of g.

$\frac{\left|a\right|}{\left|g\right|}=\frac{40.0\:m/s^2}{9.80\:m/s^2}$

$\frac{\left|a\right|}{\left|g\right|}=4.08$

Therefore,

$a=4.08g$

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