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A roller coaster is traveling at 13m/s when it approaches a hill that is 400 m. long. Heading down the hill, it accelerates 4.0m/s^2 . What is the final velocity of the roller coaster?

Answer :

skyluke89

Answer:

58.0 m/s

Explanation:

We can solve the problem by using the following SUVAT equation:

[tex]v^2 = u^2 +2as[/tex]

where

v is the final velocity of the roller coaster

u = 13 m/s is the initial velocity

a = 4.0 m/s^2 is the acceleration

s = 400 m is the displacement

Substituting and solving for v, we find

[tex]v=\sqrt{u^2 +2as}=\sqrt{(13)^2 +2(4)(400)}=58.0 m/s[/tex]

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