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.

Velocity's rate of change with time, both in terms of speed and direction. A point or object going straight ahead is accelerated when it increases or decelerates. Even if the speed remains constant, motion on a circle increases because the direction is always shifting.

Briefing:

We utilise the relations and kinematics for this exercise.

The train accelerates in the first section from v0 = 7.0 m/s to 13 m/s in a time of t = 8.0 s.

v = v₀ + a t

a = [tex]\frac{v-v_o}{t}[/tex]

a = 13-7/8

a = 0.75 m / s²

Second part Accelerate again for t = 16 seconds

v = v₁ + a t

For this interval the initial velocity is v₁ = 13 m / s

v = 13 + 0.75 16

v = 25 m / s

According to the given statement the final speed is v = 25 m / s.

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