Question

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth
to the moon, a distance of 384,000 km. Suppose it accelerates at 20.0 m/s2 for the
first 15.0 min. of the trip, then travels at constant speed until the last 15.0 min. when
it accelerates at –20 m/s2, just coming to rest as it reaches the moon.
(a) What is the maximum speed attained?
(b) What fraction of the total distance is traveled at constant speed?
(c) What total time is required for the trip?

Answer 1

We will divide the total journey in 3 parts :

The first part is:

$$ \begin{gathered} a_{1}=20 \frac{m}{s^{2}} \\ t_{1}=15 \mathrm{~min}=900 \mathrm{~s} \end{gathered} $$

\(U\) sing the equation:

$$ \begin{gathered} v_{f_{1}}=v_{i_{1}}+a_{1} t_{1} \quad \text { in this case } \rightarrow v_{i_{1}}=0 \\ v_{f_{1}}=20 \frac{m}{s^{2}}(900 s)=18,000 \frac{m}{s} \end{gathered} $$

This is the maximum speed of the spaceship.

The distance is:

$$ x_{f_{1}}=\frac{1}{2} a_{1} t^{2}=8.1 \times 10^{6} \mathrm{~m} $$

The second part is :

$$ \begin{gathered} a_{2}=-20 \frac{m}{s^{2}} \\ t_{2}=15 \min =900 s \\ v_{f_{2}}=v_{i_{3}}=-18,000 \frac{m}{s} \end{gathered} $$

The third part is when the velocity is constant:

To know the distance traveled on this part, we apply the next equation :

$$ x_{f_{3}}-x_{i_{3}}=\frac{1}{2}\left(v_{1_{3}}+v_{f_{3}}\right) t \quad \rightarrow t=\frac{x_{f_{3}}}{v_{f_{1}}} $$

since \(v_{f_{1}}=v_{i_{3}}=v_{i_{2}}\)

but, \(x_{3}=d-\left(x_{1}+x_{2}\right) \rightarrow x_{1}=x_{2} \quad x_{3}=d-2\left(x_{1}\right) \quad x_{3}=367,800,00 \mathrm{~m}\)

$$ t_{3}=20,433.33 \mathrm{~s} $$

The total time is :

$$ t_{t}=t_{1}+t_{2}+t_{3} \rightarrow t_{1}=t_{2} \quad \rightarrow \quad t_{t}=2 t_{1}+t_{3}=22,233.33 \mathrm{~s} $$