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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 4.0 m/s . Once free of this area, it speeds up to 11 m/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. Part A What is the final speed

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Answer:

25 m/s

Explanation:

Given that:

Initial speed, u = 4 m/s

Final velocity, V = 11 m/s

Time, t = 8 seconds

t2, = 16 seconds

Acceleration, a= (change in velocity) / time interval

a = (11 - 4) / 8

a = 7 / 8 = 0.875m/s²

Final velocity, v2 ;

Acceleration * t2

0.875 * 16 = 14

V2 = 14 m/s

Final speed : v + v2 = (11 + 14)m/s = 25m/s

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