The final velocity of the roller coaster is 58.0 m/s
Given data:
The initial speed of roller coaster is, u = 13 m/s.
The path length of hill is, L = 400 m.
The acceleration of coaster is, [tex]a=4.0 \;\rm m/s^{2}[/tex].
Since, the coaster is approaching to hill at variable speed. Then, apply the third kinematic equation of motion to obtain the final velocity (v) as,
[tex]v^{2}=u^{2}+2aL[/tex]
Substitute the values as,
[tex]v^{2}=13^{2}+(2 \times 4 \times 400)\\v=\sqrt{13^{2}+(2 \times 4 \times 400)} \\v \approx58.0 \;\rm m/s[/tex]
Thus, we can conclude that the final velocity of the roller coaster is 58.0 m/s.
Learn more about the kinematic equations here:
https://brainly.com/question/14355103
