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1 year ago

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As a train accelerates away from a station, it reaches a speed of 4.9 m/s in 5.4 s. if the train's acceleration remains constant

, what is its speed after an additional 5.0 s has elapsed?

1 answer:

romanna [79]1 year ago

3 0

The speed of the train after an additional 5 s has elapsed is 9.45 m/s

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

a is the acceleration
v is the final velocity
u is the initial velocity
t is the time

<h3>How to determine the acceleration in 5.4 s</h3>

Initial velocity (u) = 0 m/s
Final velocity (v) = 4.9 m/s
Time (t) = 5.4 s
Acceleration (a) =?

a = (v – u) / t

a = (4.9 – 0) / 5.4

a = 4.9 / 5.4

a = 0.91 m/s²

<h3>How to determine the final velocity after 5 s</h3>

Initial velocity (u) = 4.9 m/s
Acceleration (a) = 0.91 m/s²
Time (t) = 5 s
Final velocity (v) = ?

a = (v – u) / t

0.91 = (v – 4.9) / 5

Cross multiply

v – 4.9 = 0.91 × 5

v – 4.9 = 4.55

Collect like terms

v = 4.55 + 4.9

v = 9.45 m/s

Learn more about acceleration:

brainly.com/question/491732

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$\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825$

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The pressures at $s_2 \ and\ s_3$ is:

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