91Ó°ÊÓ

• The initial velocity of the rocket;$u=0 m/s$.
• The final velocity of the rocket;$v=224 m/s$.
• Time taken by the rocket to achieve the final speed; $t=0.900 s$.

Newton’s first law of motion can be expressed as,

$v=u+at$ ……………………… (1)

Here v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Substituting the given data in the above expression, we get

$$\begin{gathered} 224\mathrm{m}/\mathrm{s}=0\mathrm{m}/\mathrm{s}+a×0.900\mathrm{s} \\ \mathrm{a}=\frac{224\mathrm{m}/\mathrm{s}}{0.900\mathrm{s}} \\ \mathrm{a}=248.9\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration of the rocket is$248.9 \mathrm{m}/{\mathrm{s}}^2$

The acceleration of a freely falling body is $g=9.8 m/s^2$.

Therefore, the ratio between rocket acceleration and acceleration due to gravity is,

$$\begin{gathered} \frac{a}{g}=\frac{248.9\mathrm{m}/{\mathrm{s}}^2}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =25.39 \end{gathered}$$

Thus, the ratio between rocket acceleration and acceleration due to gravity is 25.39.

Newton’s first law of motion can be expressed as,

$\mathbf{d}\mathbf{=}\mathbf{ut}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{at}}^{\mathbf{2}}$ ………………………… (2)

d is the distance traveled.

Substituting the given data in the above expression, we get:

$$\begin{gathered} d=(0\mathrm{m}/\mathrm{s})×\left(0.900\mathrm{s}\right)+\frac{1}{2}\left(248.9\mathrm{m}/{\mathrm{s}}^2\right)(0.900\mathrm{s}{)}^2 \\ =100.80\mathrm{m} \end{gathered}$$

Thus, the distance covered in 0.900 s is 100.80 m.

• The initial speed of the sled is 283 m/s
• The final speed of the sled is 0 m/s
• The time duration is 1.40 s

Substituting these values in the equation (1), we get

$$\begin{gathered} a=\frac{−283\mathrm{m}/\mathrm{s}}{1.40\mathrm{s}} \\ a=−202.1\mathrm{m}/{\mathrm{s}}^2 \\ \frac{\mathrm{a}}{\mathrm{g}}=\frac{−202.1\mathrm{m}/{\mathrm{s}}^2}{−9.8\mathrm{m}/{\mathrm{s}}^2} \\ \frac{\mathrm{a}}{\mathrm{g}}=20.6 \end{gathered}$$

This acceleration value is almost half of their given value of 40 g.

Thus, the figures are inconsistent.