91Ó°ÊÓ

Mass of the rocket = 4.5 million pounds

When time is t =8 s, the speed of rocket ship is,

$v=161 km/h=161 km"/h×\left(\frac{1000 m}{1 km}\right)×\left(\frac{1 hr}{3600 s}\right)=44.73 \mathrm{m}/\mathrm{s}$

When time is $\mathrm{t}=1 \min =1 \min ×\left(\frac{60 \mathrm{s}}{1\min }\right)=60 \mathrm{s}$, the speed of rocket ship is,

$v=1610 km/h=1610 km/h×\left(\frac{1000 m}{1 km}\right)×\left(\frac{1 hr}{3600 s}\right)=447.3 \mathrm{m}/\mathrm{s}$

The relation between acceleration speed and time can be given as,

$a=\frac{\mathrm{Δ}v}{\mathrm{Δ}t}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(1)

Where, $a, \mathrm{Δ}v, \mathrm{and} \mathrm{Δ}t$ are the acceleration, difference in speed, and time duration, respectively.

Since the rocket is launched from rest, therefore at t =0 speed is 0 m/s and at t =8 s, speed is 44.73 m/s, therefore,

$$\begin{gathered} \mathrm{a}=\frac{44.73\mathrm{m}/\mathrm{s}-0\mathrm{m}/\mathrm{s}}{8\mathrm{s}-0\mathrm{s}} \\ =5.59\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The average acceleration in the first 8 s is$5.59\mathrm{m}/{\mathrm{s}}^2$.

The speed at t =8 s is 44.73 m/s, and at t =60 s is 447.3 m/s, therefore, substituting the values in the equation (1),

Then we know that,

$$\begin{gathered} \mathrm{a}=\frac{447.3\mathrm{m}/\mathrm{s}-44.73\mathrm{m}/\mathrm{s}}{60\mathrm{s}-8\mathrm{s}} \\ =7.74\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The average acceleration in between 8 s to 1 min is$7.74\mathrm{m}/{\mathrm{s}}^2$.

The total distance traveled by the rocket ship can be calculated as,

$s=ut+\frac{1}{2}a{t}^2$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(2)

Where u is the initial speed, a is the acceleration and t is the travel time.

Substituting u=0 m/s $\mathrm{a}=5.59\mathrm{m}/{\mathrm{s}}^2$and t =8 s in the equation (2),

$$\begin{gathered} \mathrm{s}=\left(0\mathrm{m}/\mathrm{s}\right)×\left(8\mathrm{s}\right)×\frac{1}{2}×5.59\mathrm{m}/{\mathrm{s}}^2×{\left(8\mathrm{s}\right)}^2 \\ =178.88\mathrm{m} \end{gathered}$$

Thus, the distance traveled in the first 8 s is 178.88 m.

Substituting u =44.73m/s $\mathrm{a}=7.74\mathrm{m}/{\mathrm{s}}^2$and t =52 s in the equation (2), we get,

$$\begin{gathered} \mathrm{s}=\left(44.73\mathrm{m}/\mathrm{s}\right)×\left(52\mathrm{s}\right)×\frac{1}{2}×7.74\mathrm{m}/{\mathrm{s}}^2×{\left(52\mathrm{s}\right)}^2 \\ =2325.96\mathrm{m}+10464.48\mathrm{m} \\ =12790.44\mathrm{m} \end{gathered}$$

Thus, the total distance traveled between the 8 s to 1 min is $12790.44\mathrm{m}$.