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?

Question

Bentley Clay

Physics

9 January, 11:31

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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Answers (1)

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Cordell Blackburn · Answer 1

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