Question

The bus, at location A, is traveling at 30 m/s when it is shifted to neutral and allowed to

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity of the bus at B is $v_B= 31.6\ m/s$

Explanation:

Let's take  position B as base point .

From the diagram  height between point B and A ia mathematically evaluated as

        $h_A = 60 -55$

      $h_A = 5 \ m$

From the question we are told that

   The velocity at location A is  $v_A = 30 m/s$

       

     According the law of conservation of energy

          $PE_A + KE_A = KE_B$

Where $PE_A$ is the potential energy at A which is mathematically represented as

          $PE_A = mgh_A$

$KE_A$ is the kinetic energy energy at A which is mathematically represented as

            $KE_A = \frac{1}{2} mv^2_A$

$KE_B$ is the  kinetic energy at A which is mathematically represented as

          $KE_B = \frac{1}{2} mv_B^2$

Where $v_B$ is the velocity at location B

So

      $mgh_A + \frac{1}{2} mv_A^2 = \frac{1}{2} mv_B^2$

Making $v_B$ the subject of the formula

       $v_B= \sqrt{\frac{gh_A + \frac{1}{2} v_A^2 }{0.5} }$

Substituting values

       $v_B= \sqrt{\frac{(9.8 * 5) + \frac{1}{2} 30^2 }{0.5} }$

        $v_B= 31.6\ m/s$