2.104 Alternative Exercise 2.104 A spaceship ferrying workers to Moon Base l tak

Question

2.104

Alternative Exercise 2.104 A spaceship ferrying workers to Moon Base l takes a straight-line path from the earth to the moon, a distance of 384000 km. Suppose it accelerates at an acceleration 21.0 m/s2 for the first time interval 16.0 min of the trip, then travels at constant speed until the last time interval 16.0 min hen it accelerates at 21.0 m/s just coming to rest as it reaches the moon. What fraction of the total distance is traveled at constant speed? 19350.912 Submit My Answers Give Up Incorrect, Try Again, 2 attempts remaining Part C What total time is required for the trip? t 2880 Submit My Answers Give Up Incorrect, Try Again, 4 attempts remaining

Explanation / Answer

The appropriate distance formula is
s = ut + 0.5at^2 where
u = 0, t = 16*60, a = 21, which gives
s =9676800. This is metre, so the distance covered during acceleration is
9676800 km.
Since the rate of deceleration at the end is the same (21m/s^2), the total distance covered in accelerating and decelerating is
193536 km

so this is the maximum speed, and the speed at which the craft travels the remaining
190 464 km.

(a) The speed reached is given by
v = u + at, which is
21*960
= 20 160 m/s

(b) Fraction of total distance
= 190464/384000
= 0.496

(c) Hence time travelled at constant speed
= dist/speed
= 21243.3333 seconds.

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