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a woodpecker’s brain is specially protected from large accelerations 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 2.00 mm. (a) find the acceleration in meters per second squared and in multiples of g, where g



Answer :

The acceleration in meters per second squared and in multiples of g is

a = 9.18 g.  

Acceleration is the rate of change of velocity with respect to time.

"comes to a stop" means

final v = 0 m/s

⇒v=0m/s.

"from an initial velocity of"

u = 0.600 m/s

"in a distance of 2.00 mm"

x  = 2.00 × (10)⁻³m

So, we need an equation that involves

a, v, u,  and x,

the equation is —

v²=(u)²-2×a×s

0= (0.600)² – 2×a× 2.00 ×(10)⁻³​

In solving this, we get a;

a=90 m/s² —>  eq. 1

Now, find the acceleration in meters per second squared and in multiples of g.

And, we know that g=9.8 m/s²2. —> eq.2

Divide eq.1 by eq.2

We get,

= 9.18 ⇒ a = 9.18 g.  

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