contestada

015 10.0 points
A subway train starting from rest leaves a
station with a constant acceleration. At the
end of 7.13 s, it is moving at 12.4062 m/s.
What is the train's displacement in the first
4.91257 s of motion?
Answer in units of m.

Respuesta :

Answer:

20.996 m

Explanation:

Given:

Initial velocity, [tex]u=0 \textrm{ m/s}[/tex]

Final velocity, [tex]v=12.4062 \textrm{ m/s}[/tex]

Total time taken, [tex]t_{Total} = 7.13 [/tex] s.

∴ Acceleration is given as,

[tex]a=\frac{v-u}{t_{Total}}=\frac{12.4062-0}{7.13}=1.74[/tex] m/s²

Now, using Newton's equation of motion, we find the displacement.

Displacement is given as:

[tex]s=ut+\frac{1}{2} at^{2}[/tex]

Plug in 0 for [tex]u[/tex], 4.91257 for [tex]t[/tex] and 1.74 for [tex]a[/tex]. Solve for [tex]s[/tex].

This gives,

[tex]s=0+\frac{1}{2} \times 1.74 \times (4.91257)^{2}=20.996 \textrm{ m}[/tex]

Therefore, the train's displacement in the first 4.91257 s of motion is 20.996 m.

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