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.