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Andrew [12]
2 years ago
10

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 . Sim

ultaneously, 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
Physics
1 answer:
Nonamiya [84]2 years ago
8 0

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

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