91Ó°ÊÓ

• 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.

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

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,

$$\begin{gathered} (0{)}^2-(0.6{)}^2=2×\left(a\right)×(0.002) \\ -0.36=2×(0.002)×a \\ a=\frac{-0.36}{2×(0.002)} \\ a=-90{\text{m/s}}^{\text{2}} \end{gathered}$$

Hence the deceleration of the woodpecker is -90 m/s².

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,

$$\begin{gathered} 0=0.6+(-90)×t \\ t=\frac{0.6}{90} \\ t=0.006\text{s} \end{gathered}$$

Hence the time of stopping is 0.006 s.

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

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,

$$\begin{gathered} (0{)}^2-(0.6{)}^2=2×\left(a\right)×(0.0045) \\ -0.36=2×(0.0045)×a \\ a=\frac{-0.36}{2(0.0045)} \\ a=-40{\text{m/s}}^{\text{2}} \end{gathered}$$

Hence the Deceleration of the substance is -40 m/s².

For the deceleration in multiple of g, we have to divide the deceleration by9.81.

The deceleration in multiple of g:

$$\begin{gathered} a=\frac{-40\text{g}}{9.81} \\ \text{a}=-4.077\text{g} \end{gathered}$$

Hence the deceleration is -40 m/s² and in multiple of g is -4.077 g.