Question

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?

Answers

  1. Answer:

    The taken is  t_A = 19.0 \ s

    Explanation:

    Frm the question we are told that

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

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

         The distance of car B  from A is  d = 300 \ m

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

    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  

              d = v_B * t_A

    Where t_B is the time taken by car B

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

          d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300

    Now substituting values

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

    Equating the both d

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

    substituting values

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

       7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

      7 t_A =1.2 t_A^2 - 300

       1.2 t_A^2 - 7 t_A - 300 = 0

    Solving this using quadratic formula we have that

         t_A = 19.0 \ s

Leave a Comment