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?

Answer 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

     $t_A = 19.0 \ s$