The head of a woodpecker is moving forward at a speed of 7.49 m/s when the beak makes first contact with a tree limb. The beak stops after penetrating the limb by 1.87 mm. Assuming the acceleration to be constant, find the acceleration magnitude in terms of g.
Know: velocity (v) = 0, initial velocity (v0) = 7.49m/s, and displacement during acceleration (x-x0) is 1.87x10^-3m.
v^2 = v0^2 + 2a(x-x0)
The Attempt at a Solution
I was given the solution, but do not understand where a certain figure comes from. They plug the known quantities into the equation to get 0^2 = 7.49^2 + 2a(1.87*10^-3m), which algebraically reduce to a = -1.500 x 10^4m/s^2.
However, the next step loses me. They divide by g = 9.8m/s^2, take the absolute value, and that gives the magnitude of the head's acceleration is a = (1.53x10^3).
Where does that 9.8m/s^2 come from?! Thanks!