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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 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 16s to reach a higher cruising speed.

Required:
What is the final speed?

Respuesta :

Answer:

 v = 25 m / s

Explanation:

For this exercise we use the relations and kinematics

first part, the train accelerates from v₀ = 7.0 m/s to 13 m/s in a time t = 8.0 s

        v = v₀ + a t

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

        a = [tex]\frac{13-7}{8}[/tex]

        a = 0.75 m / s²

second part. Accelerate again for t = 16 s

         v = v₁ + a t

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

         v = 13 + 0.75  16

         v = 25 m / s

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