Question

A roller coaster is traveling at 13 m/s when it approaches a hill that is 400 m long. Heading down the hill, it accelerates at 4.0 m/s2. What is the final velocity of the roller coaster?

Answer 1

Answer:58 m/s

Step-by-step explanation:

That’s what it says

Answer 2

The value is
`v = 47 \  m/s`

Step-by-step explanation:

From the question we are told that

The initial speed of the roller coaster is
`u =  13 \  m/s`

The length of the hill is
`l   = 400 \  m`

The acceleration of the roller coaster is
`a=4.0 \ m/s^2`

Generally the acceleration is mathematically represented as

`a =  ( v - u)/( t_f -  t_i )`

Here
`t_i` is the initial time which is equal to zero

`v_f` is the final velocity which is mathematically represented as

`v_f  =  (d)/( t_f)`

So

`a =  ( (d)/(d_f)  - u )/( t_f - t_i)`

`4 = ((400)/( t_f)  - 13)/(t_f - 0)`

`4 =  (400 - 13t_f)/( t_f) *  (1)/(t_f)`

`4t_f ^2  +13f  + 400 =`

Solving this using quadratic formula we obtain

`t_f =  8.5 \ s`

`t_f =  -11.8 \ s`

Generally time cannot be negative so

`t_f =  8.5 \ s`

Generally the final velocity is mathematically represented as

`v = (400)/(8.5)`

`v = 47 \  m/s`