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olga2289
17 days ago
6

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 7.0m/s . Once free of this area, it speeds up to 12m/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 Speed?
Physics
1 answer:
Yuliya22 [1.1K]17 days ago
5 0

Answer:

v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s

Explanation:

In this scenario, we determine the initial velocity as follows:

v_i = 7 \frac{m}{s}

The final velocity in this instance can be expressed as:

v_f = 13 \frac{m}{s}

It is noted that transitioning from 7m/s to 13m/s takes 8 seconds. We can apply a specific kinematic equation to find the acceleration for the first part of the journey:

v_f = v_i +at

Solved for acceleration, we find:

a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}

For the subsequent route, we assume constant acceleration and that the train continues for 16 seconds, beginning with an initial velocity of 13m/s from the previous segment, allowing us to calculate the final speed via the following formula:

v_f = v_ i +a t

Substituting into the equation yields:

v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s

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