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3 years ago

14

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 4.0 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. Part A What is the final speed

1 answer:

Nimfa-mama [501]3 years ago

3 0

Answer:

25 m/s

Explanation:

Given that:

Initial speed, u = 4 m/s

Final velocity, V = 11 m/s

Time, t = 8 seconds

t2, = 16 seconds

Acceleration, a= (change in velocity) / time interval

a = (11 - 4) / 8

a = 7 / 8 = 0.875m/s²

Final velocity, v2 ;

Acceleration * t2

0.875 * 16 = 14

V2 = 14 m/s

Final speed : v + v2 = (11 + 14)m/s = 25m/s

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Answers:

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$V=V_{o}+gt$ (2)

Where:

$y=0$ is the final height of the steel ball

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$V_{o}=0$ is the initial velocity of the steel ball (it was dropped)

$V$ is the final velocity of the steel ball

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<u>Knowing this, let's begin with the answers:</u>

<h2>a) Time it takes the steel ball to reach the ground</h2>

We will use equation (1) with the conditions listed above:

$0=y_{o}+\frac{1}{2}gt^{2}$ (3)

Isolating $t$ :

$t=\sqrt{\frac{-2y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2(400 m)}{-9.8 m/s^{2}}}$ (5)

$t=9.035 s$ (6)

<h2>b) Final velocity of the steel ball</h2>

We will use equation (2) with the conditions explained above and the calculaated time:

$V=gt$ (7)

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$V=-88.543 m/s$ (9) The negative sign indicates the direction of the velocity is downwards

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$\lambda=\frac{\lambda_0}{n}$

where

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