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:

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

         $v = \frac{400}{8.5}$

         $v = 47 \ m/s$