Aspaceship ferrying workers to moon base i takes a straight-line path from the earth to the moon, a distance of 384000 km. suppose it accelerates at an acceleration 19.9 m/s2 for the first time interval 14.7 min of the trip, then travels at constant speed until the last time interval 14.7 min , when it accelerates at − 19.9 m/s2 , just coming to rest as it reaches the moon. part a what is the maximum speed attained?

You have to assume something about the starting speed. I will take it to be 0 although you know that isn't so.

t = 14.7 min = 14.7 min *[ 60 s / 1 min] = 882 seconds.
vi = 0 m/s
a = 19.9 m/s^2
vf = ???

Formula

a = (vf - vi)/t

Substitute and solve

19.9 = (vf - 0)/882 
19.9 * 882 = vf
vf = 17552 m/s

That seems large. However the time is very long, so you should expect the final velocity to be large.

From your givens, you have three sig digs so the answer is
1.76 * 10^4 m/s

 

Did you ever figue this

B) as the temperature increases, the kinetic energy of the particles increases.|-/