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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

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Answer:

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

Step-by-step 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 = (z)/(v_A)

The distance covered by puck B is mathematically represented as


18 - z = v_B * t

=>
t = (18 - z)/(v_B)

Since the time take before collision is the same


(18 - z)/(V_B) = (z)/(v_A)

substituting values


(18 -z )/(4.3) = (z)/(3.90)

=>
70.2 - 3.90 z = 4.3 z

=>
z = 8.56 \ m

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