Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with a uniform forward acceleration of 2.40 m/s2. How long after car A begins to accelerate does it take car A to overtake car B?

Respuesta :

Answer:

The taken is  [tex]t_A = 19.0 \ s[/tex]

Explanation:

Frm the question we are told that

  The speed of car A is  [tex]v_A = 22 \ m/s[/tex]

   The speed of car B is  [tex]v_B = 29.0 \ m/s[/tex]

     The distance of car B  from A is  [tex]d = 300 \ m[/tex]

     The acceleration of car A is  [tex]a_A = 2.40 \ m/s^2[/tex]

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          [tex]d = v_B * t_A[/tex]

Where [tex]t_B[/tex] is the time taken by car B

Now this can also be represented as using equation of motion as

      [tex]d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300[/tex]

Now substituting values

       [tex]d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]

Equating the both d

       [tex]v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]

substituting values

   [tex]29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]

   [tex]7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]

  [tex]7 t_A =1.2 t_A^2 - 300[/tex]

   [tex]1.2 t_A^2 - 7 t_A - 300 = 0[/tex]

Solving this using quadratic formula we have that

     [tex]t_A = 19.0 \ s[/tex]

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