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A train starts from rest (at position zero) and moves with constant acceleration. On the first observation, its velocity is 20m/s and 80seconds later the velocity became 60m/s. At 80s calculate the position, average velocity, and the constant acceleration over the interval.(7-points)

Sagot :

whitneytr12

The value of the acceleration is a = 0.5 m/s². The position at 80 s is x = 3200 m and finally the average velocity is v = 40 m/s.

Acceleration:

We can use the fallowing kinematic equation to get the acceleration at 80 s.

[tex]a=\frac{v_{f}-v_{i}}{t}[/tex]            

Where:

  • v(i) is the initial velocity (20 m/s)
  • v(f) is the final velocity (60 m/s)
  • t is the interval (80 s)

The, we have:

[tex]\vec{a}=\frac{60-20}{80}[/tex]

[tex]\vec{a}=0.5\: m/s^{2}[/tex]

Position:

Knowing the acceleration we can find the position using the falling equation.

[tex]\vec{x}=v_{i}t+0.5at^{2}[/tex]

[tex]\vec{x}=20*80+0.5*0.5*80^{2}[/tex]

[tex]\vec{x}=3200 m[/tex]

Average velocity:

The definition of the average velocity is:

[tex]\vec{v}=\frac{\Delta x}{t}[/tex]

[tex]\vec{v}=\frac{x_{f}-x_{i}}{t}[/tex]

[tex]\vec{v}=\frac{3200-0}{80}[/tex]

[tex]\vec{v}=40\: m/s[/tex]

Learn more about the kinematic equations here:

https://brainly.com/question/13143668

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