Alight-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s2 for 20 s. it then proceeds at constant speed for 1100 m before slowing down at 2.2 m/s2 until it stops at the station.
how much time does it take the train to go between the stations?
express your answer using two significant figures. the answer is not 20s, as was answered and "verified" previously.

It accelerates from rest at 1.1 m/s^2 for 20s. 

s = at^2/2 
s = (.5)(1.1)(20)^2= 220m 

then it travels 1100m 

Then it slows down at 2.2m/s^2 until it stops at the station. 

Calculate the train's speed: 
v = at 
v = (1.1)(20) = 22m/s 

Now, calculate the time to stop: 
v = at 
t = 22/2.2 = 10s 
s = t(u+v)/2 
s = 10(0+22)/2 = 110m 

(A) Total distance = 220 +1100+110 = 1430m 

(B) Time = 20 + 1100/22 + 10 = 80s

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.