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 . Simultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 1.30 , and puck B moves with a speed of 1.70 .

What is the distance covered by puck A by the time the two pucks collide?

Answer 1

Puck A covers a distance of 7.8 meters before it collides with puck B, which is calculated using the relative speed at which the pucks approach each other and the time taken for them to collide.

Step-by-step explanation:

To find the distance covered by puck A by the time the two pucks collide, we need to consider the relative velocity at which they are approaching each other. Since puck A moves with a speed of 1.30 meters per second and puck B moves with a speed of 1.70 meters per second, they are closing in on each other at a velocity of 1.30 m/s + 1.70 m/s = 3.00 m/s. The initial separation between the two pucks is 18.0 meters.

The time it takes for the pucks to collide can be calculated using the formula: time = distance / relative speed. Plugging in the values, we get time = 18.0 m / 3.00 m/s, which simplifies to 6.0 seconds.

To determine the distance covered by puck A, we multiply its velocity by the time taken to collide: distance A = velocity A x time. Therefore, distance A = 1.30 m/s x 6.0 s, resulting in puck A covering a distance of 7.8 meters.

Answer 2

The pucks are traveling towards one another, so their relative velocity becomes:
1.3 + 1.7 = 3 m/s
The time it will take for them to collide:
18/3 = 6 seconds

In this time, the distance each puck will cover:
Da = 1.3 x 6
= 7.8 m

Db = 1.7 x 6
= 10.2 m