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 an acceleration 20.2 m/s2 for the first time interval 15.8 min of the trip, then travels at constant speed until the last time interval 15.8 min , when it accelerates at − 20.2 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

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

$15.8 min*\frac{60s}{1min}=948s$

The equation of motion you need is:

$v_f = a*t+v_0$

where $v_f$ is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

$v_f = a*t = 20.2*948 = 19,149.6 m/s$

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

$x=\frac{v_0+v_f}{2}t$

solving for first and last interval:

Since the spaceship starts and finish with 0 velocity:

$x=\frac{v}{2}t=\frac{19,149.6}{2}948=9,076,910.4m=9,076.9104km$

Then the ship traveled $384,000-9,076.9104*2 = 361,846.1792km$ at constant speed, which means that it traveled:

$f_{constant_speed} =\frac{ x_{constant_speed}}{x_total} =\frac{361,846.1792}{380,000} =0.95$

Which in percentage is 95% of the trip.

to calculate total time you need to calculate the time used during constant speed:

$t = \frac{361,846,179.2}{19,149.6} = 18,895.75s = 314min$

That added to the other interval times:

$t_{total} = t_1+t_2+t_3=15.8+314.93+15.8=346.5min$