A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon?

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

The appropriate distance formula is
s = ut + 0.5at^2 where
u = 0, t = 15*60, a = 20, which gives
s = 810 000. This is metre, so the distance covered during acceleration is
810 km.
Since the rate of deceleration at the end is the same (20m/s^2), the total distance covered in accelerating and decelerating is
1620 km

(a) The speed reached is given by
v = u + at, which is
20*900
= 18 000 m/s
=18 km/s,
so this is the maximum speed, and the speed at which the craft travels the remaining
382 380 km.
(b) Fraction of total distance
= 382380/384000
= 6373/6400

(c) Hence time travelled at constant speed
= dist/speed
= 21243.3333 seconds.
= 5hour 54min 3.33... sec
For total time, add the 15 minutes at each end,
getting
6hour 24min 3.33... sec

Did I make a mistake? Sure the method's correct, but check my working!

Answer 2

final velocity 18 km/s. total time at a constant velocity is roughly 6 hours, which is 91 % of total traveling time.

Answer 3

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