91Ó°ÊÓ

Launch distance,=$5.0\mathrm{μ}m$

The speed in launch is 1.6 m/s

The distance for which speed reduces to zero is 1.0 mm

The value of acceleration or deceleration can be found using the kinematic equations. The acceleration would be positive in sign whereas deceleration would be negative in sign.However, one needs to be careful about the directions of initial velocity and final velocity as they can cause the signs of acceleration to change accordingly.

The formula is as given below,

$v_f^2=v_i^2+2ax$ (i)

Here,$v_f$ is the final speed,$v_i$ is the initial speed, a is the acceleration and x is the distance.

Convert $5.0\mathrm{μ³¾}$to meter.

$$\begin{gathered} 5.0\mathrm{μ³¾}=5.0\mathrm{μ³¾}×\frac{1×{10}^{-6}\mathrm{m}}{1.0\mathrm{μ³¾}} \\ =5×{10}^{-6}\mathrm{m} \end{gathered}$$

Putting values in equation (i),

$$\begin{gathered} 1.6^2=0^2+2×a×5×{10}^{-6} \\ a=\frac{2.56}{{10}^{-5}} \\ =2.56×{10}^{5}m/s \end{gathered}$$

Since, the value of$\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$ therefore, the acceleration during the launch is,

$$\begin{gathered} a=2.56×{10}^5\mathrm{m}/{\mathrm{s}}^2×\frac{\mathrm{g}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =2.6×{10}^4\mathrm{g} \end{gathered}$$

Thus, the value of acceleration during the launch in terms of g is$2.6×{10}^4\mathrm{g}$ .

For the speed reduction, the initial speed is 1.6m/s and final speed is zero, and the distance is 1.0 mm .

Convert 1.0 mm to meter.

$$\begin{gathered} 1.0\mathrm{mm}=1.0\mathrm{mm}×\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}} \\ =1.0×{10}^{-3}\mathrm{m} \end{gathered}$$

Putting values in equation (i),

$$\begin{gathered} v_f^2=v_i^2+2ax \\ {0}^2=1.6^2+2a×1.0×{10}^{-3} \\ a=-1.28×{10}^3\mathrm{m}/\mathrm{s} \end{gathered}$$

Divide the acceleration by the value of g to get the acceleration in terms of g.

$$\begin{gathered} a=-1.28×{10}^3\mathrm{m}/{\mathrm{s}}^2×\frac{\mathrm{g}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =-1.30×{10}^2\mathrm{g} \end{gathered}$$

Thus, the acceleration in terms of g during the speed reduction is$-1.30×{10}^2\mathrm{g}$.