Question

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Simultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 3.90 m/s , and puck B moves with a speed of 4.30 m/s . What is the distance covered by puck A by the time the two pucks collide

Answer 1

Answer:

The distance covered by puck A before collision is  $z = 8.56 \ m$

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  $v_A = 3.90 \ m/s$

        The speed of puck B is  $v_B = 4.30 \ m/s$

The distance covered by puck A is mathematically represented as

     $z = v_A * t$

  =>  $t = \frac{z}{v_A}$

 The distance covered by puck B  is  mathematically represented as

      $18 - z = v_B * t$

=>   $t = \frac{18 - z}{v_B}$

Since the time take before collision is the same

        $\frac{18 - z}{V_B} = \frac{z}{v_A}$

substituting values

          $\frac{18 -z }{4.3} = \frac{z}{3.90}$

=>      $70.2 - 3.90 z = 4.3 z$

=>       $z = 8.56 \ m$