Q29E - University Physics with Modern Physics

Question

At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h; at the end of the first 1.00 min, its speed is 1610 km/h. (a) What is the average acceleration (in m/s²) of the rocket (i) during the first 8.00 s and (ii) between 8.00 s and the end of the first 1.00 min? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the rocket travel (i) during the first 8.00 s and (ii) during the interval from 8.00 s to 1.00 min?

Step-by-Step Solution

Answer

(a) Average acceleration

 

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

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

(b) Distance traveled

 

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

(ii) Thus, the total distance traveled between the 8 s to 1 min is 12790.44 m.

Step 1: Identification of the given data

Mass of the rocket = 4.5 million pounds

 

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

$v=161\hspace{0.33em}km /h=161\hspace{0.33em}km"/h\times \left(\frac{1000\hspace{0.33em}m}{1\hspace{0.33em}km}\right) \times \left(\frac{1\hspace{0.33em}hr }{3600\hspace{0.33em}s}\right)=44.73\hspace{0.33em}\mathrm{m}/\mathrm{s}$

When time is $\mathrm{t}=1\hspace{0.33em}\min = 1\hspace{0.33em}\min \times \left(\frac{60\hspace{0.33em}\mathrm{s}}{1 \min }\right)=60\hspace{0.33em}\mathrm{s}$, the speed of rocket ship is,

$v=1610\hspace{0.33em}km /h=1610\hspace{0.33em}km/h\times \left(\frac{1000\hspace{0.33em}m}{1\hspace{0.33em}km}\right) \times \left(\frac{1\hspace{0.33em}hr }{3600\hspace{0.33em}s}\right)=447.3\hspace{0.33em}\mathrm{m}/\mathrm{s}$

Step 2: Calculation for the average acceleration

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

$a=\frac{\Delta v}{\Delta t}$………………….(1)

Where, $a,\hspace{0.33em}\Delta v,\hspace{0.33em}\mathrm{and}\hspace{0.33em}\Delta t$ 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 $5.59\mathrm{m}/{\mathrm{s}}^2$.

Step 3: Calculation for the average acceleration

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 $7.74 \mathrm{m}/{\mathrm{s}}^2$.

Step 4: Calculation for the traveled distance

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

$s=ut+\frac{1}{2}a{t}^2$…………………………(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),

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

 

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

Step 5: Calculation for the distance travel

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

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

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