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18 June, 23:17

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 6.0 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. What is the final spend?

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  1. 18 June, 23:31
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    v = 21 m / s

    Explanation:

    We can solve this exercise with the kinematics equations, let's start by finding the acceleration of the train with the initial data

    v = v₀ + a t

    the initial speed is the speed within the city 6 m / s, the final speed is v = 11 m / s and the time is t = 8 s

    a = (v-v₀) / t

    a = (11 - 6) / 8

    a = 0.625 m / s²

    when it leaves the city with speed vo = 11 m / s it accelerates for t = 16 s

    v = v₀ + a t

    v = 11 + 0.625 16

    v = 21 m / s
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