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IrinaK [193]
3 years ago
5

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?
Physics
1 answer:
SIZIF [17.4K]3 years ago
5 0

Answer:

The  value is    v = 47 \  m/s

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 =  \frac{ 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  =  \frac{d}{ t_f}

So  

     a =  \frac{ \frac{d}{d_f}  - u }{ t_f - t_i}

     4 = \frac{\frac{400}{ t_f}  - 13}{t_f - 0}

      4 =  \frac{400 - 13t_f}{ t_f} *  \frac{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 = \frac{400}{8.5}

         v = 47 \  m/s

       

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