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Initially stationary, a train has a constant acceleration of 0.8 m/s2. (a) What is its speed after 27 s? m/s (b) What is the total time required for the train to reach a speed of 41 m/s?

Respuesta :

mathmate

Answer:

(a) v1 = 21.6 m/s

(b) t = 51.25 s

Explanation:

Use kinematics equation

v1 = v0 + at

Given

v0 = 0 = initial velocity

a = 0.8 m/s^2 = acceleration

(a) t = 27 seconds

v1 = v0 + at = 0 + 0.8*27 = 21.6 m/s

(b) v1 = 41 m/s

v1 = v0 + at

solve for t

t = (v1-v0)/a = (41-0)/0.8 = 51.25 s

Cricetus

(a) The speed will be "21.6 m/s".

(b) Total time required will be "51.25 s".

Given:

  • Acceleration, a = 0.8 m/s²
  • Time, t = 41 m/s
  • Speed, v = 41 m/s

By using Kinematics equation, we get

→ [tex]v_1 = v_0+at[/tex]

(a)

At t = 27 seconds,

→ [tex]v_1 = v_0 +at[/tex]

       [tex]= 0+0.8\times 27[/tex]

       [tex]= 21.6 \ m/s[/tex]

(b)

When [tex]v_1 = 41 \ m/s[/tex]  

→ [tex]t = \frac{(v_1-v_0)}{a}[/tex]

     [tex]= \frac{41.0}{0.8}[/tex]

     [tex]= 51.25 \ s[/tex]

Thus the response above is right.

Learn more about acceleration here:

https://brainly.com/question/18401853

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