# College Physics by Openstax, Problem 2.35

#### (b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

SOLUTION:

Part a

We are given

$v_0=0\:m/s$

$a=-9.8\:m/s^2$

$x=-3.0\:m$

Note that the acceleration is due to gravity and its value is constant at $a=-9.8\:m/s^2$. Also, the distance, x, is negative because of the direction of motion. For free fall, downward motion is considered negative. To solve for the velocity just before it hits the ground, we have

$v^2=v_0^2+2ax$

$v^2=\left(0\:m/s\right)^2+2\left(-9.8\:m/s^2\right)\left(-3.0\:m\right)$

$v=\sqrt{2\left(-9.8\:m/s^2\right)\left(-3.0\:m\right)}$

$v=7.7\:m/s$

Part b

We are given

$v=0\:m/s$

$v_0=7.7\:m/s$

$x=0.02\:m$

To solve for the acceleration, we have

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

$a=\frac{\left(0\:m/s\right)^2-\left(7.7\:m/s\right)^2}{2\left(0.05\:m\right)}$

$a=-1.5\times 10^3\:m/s^2$

This is approximately 3 times the deceleration of the pilots from who were falling from thousands of meters high.

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