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 7.0 m/s . once free of this area, it speeds up to 13 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?

25 m/s

Explanation:

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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

volts times amps 6 x 0.5 3w

ok

explanation: