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madam [21]
4 years ago
15

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?
Physics
1 answer:
Reika [66]4 years ago
4 0

Answer:

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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6. Decelerating a plane at a uniform rate of -8 m/s2, a pilot stops the plane in 484 m. How
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If we are talking about an acceleration at a uniform rate, we are dealing with constant acceleration, hence we can use the following equation:

{V_{f}}^{2}={V_{o}}^{2}+2ad (1)

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V_{o} Is the initial velocity of the plane

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