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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 7.0m/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?

Answer :

Answer:

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

Explanation:

For this case we know that the initial velocity is given by:

[tex] v_i = 7 \frac{m}{s}[/tex]

The final velocity on this case is given by:

[tex] v_f = 13 \frac{m}{s}[/tex]

And we know that it takes 8 seconds to go from 7m/s to 13m/s. We can use the following kinematic formula in order to find the acceleration during the first interval:

[tex] v_f = v_i +at[/tex]

If we solve for the acceleration we got:

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

So for the other traject we assume that the acceleration is constant and the train travels for 16 s. The initial velocity on this case would be 13m/s from the first interval and we can find the final velocity with the following formula:

[tex] v_f = v_ i +a t[/tex]

And if we replace we got:

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

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