physics problem.....speed and total time?

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 20.6 m/s^2 for the first time interval 15.7 min of the trip, then travels at constant speed until the last time interval 15.7 min, when it accelerates at -20.6 m/s^2, just coming to rest as it reaches the moon.

What is the maximum speed attained?
What total time is required for the trip?

I have tried so many ways to find the correct answer but i can't seem to do it right....please help

Answer 1

Use the formual s = 0.5*a*t^2 to find the distance for each leg of the trip for which you are given the time:

s1 = 0.5*20.6*(15.7*60)^2 = 9140000 m = 9140 km

The max velocity is reached at this point, since there is no more acceleration; the formula for velocity under constant acceleration is v = a*t, so

v = 20.6*15.7*60 = 19400 m/sec = 19.4 km/sec

Since the time and acceleration for the last part are the same as the first, the last leg distance s3 is also 9140 km

The distance traveled at constant velocity is 384000 - 18280= 365720 km. The time to cover that distance is

365720 / 19.4 = 18846 sec or 314.1 min

The total trip time is then 15.7 + 314.1 + 15.7 =346 min

(I have maintained 3 sig figures in the final answers.)

Answer 2

Since both the acceleration and deceleration and times are the same, the final speed (the maximum speed} attained is
v = at = 20.6 * 15.7 x 60 = 19405.2 m/s
----------------------------------------------------------
Distance = average velocity x time.
The total distance traveled is
= (v/2) t1 + v t2 + (v/2) t3. [Note t1 = t3]
= v / 2 [2t1 +2 t2]
= v [t1 +t2] = 384,000,000 m
[t1 + t2] = 384,000,000/ 19405 = 19789 s= 329.8minutes.

Total time = [t1 +t2] + t1 [or t3] = 329.3 + 15.7
= 345.51 minutes
=5.76 hour.