By taking the formulas for KE and PE then inserting the time-dependent formulas without plugging in numbers, do algebra until t no longer appears in your answer (only constants appear such as g, Vo, m, etc.). you have to show your work but you should get energy=1/2mVo^2. energy is conserved because it is constant over all time.

25 m/s

Explanation:

from the question you van see that some detail is missing, however i found this same question using internet search engines on: 'https://www.chegg.com/homework-help/questions-and-answers/light-rail-passenger-trains-provide-transportation-within-cities-speed-slow-nearly-constan-q5808369'

here is the complete question:

'Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 7.0m/s . Once free of this area, it speeds up to 12m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. What is the final Speed?'

SOLUTION

initial speed (u) = 7 m/s

final speed (v) = 13 m/s

initial acceleration time (t1) = 8 s

final acceleration time (t2) = 16 s

what is the higher cruising speed?

acceleration =

acceleration = = 0.75 m/s^{2}

since the train accelerates at the same rate, the increase in  speed will be = acceleration x time (t2)

= 0.75 x 16 = 12 m/s

therefore the higher cruising speed = increase in speed + initial speed

= 12 + 13 = 25 m/s