After hitting with a club, the puck glides on the ice with constant acceleration. At the end of the fifth

After hitting with a club, the puck glides on the ice with constant acceleration. At the end of the fifth second after the start of the movement, its speed was equal to 1.5 m / s, and at the end of the sixth second, the puck stopped with what acceleration it was moving, what path it traveled and what was its speed at a distance of 20 meters from the start of the movement.

Given:

v (5) = 1.5 m / s – washer speed at the end of the fifth second of movement;

v (6) = 0 – puck speed at the end of the sixth second of movement;

S1 = 20 meters – distance.

It is required to determine the acceleration of the washer a (m / s ^ 2), the total distance traveled S (meter) and the speed of the washer at the distance S1 (m / s).

After the impact, the puck moved with a certain initial speed and acceleration. Then the equation of the puck speed versus time will look like this:

v (t) = v0 – a * t.

v (5) = v0 – 5 * a = 1.5 (1)

v (6) = v0 – 6 * a = 0 (2).

Subtracting from equation (1) equation (2), we get:

v0 – 5 * a – v0 + 6 * a = 1.5 – 0

a = 1.5 m / s ^ 2.

From equation (1) we find the value of the initial speed:

v0 = 1.5 + 5 * a = 1.5 + 5 * 1.5 = 1.5 + 7.5 = 9 m / s.

Then the full path traveled is:

S = v0 * t – a * t ^ 2/2 = 9 * 6 – 1.5 * 6 ^ 2/2 = 54 – 1.5 * 36/2 =

= 54 – 1.5 * 18 = 54 – 27 = 27 meters.

The puck will reach the distance S1 in the time:

t = 2 * S1 / v0 = 2 * 20/9 = 40/9 = 4.4 seconds.

Then its speed will be equal to:

v (4.4) = 9 – 1.5 * 4.4 = 9 – 6.6 = 2.4 m / s.

Answer: the acceleration of the puck is 1.5 m / s ^ 2, before stopping it will travel a distance of 27 meters, the speed of the puck after 20 meters from the start of movement will be 2.4 m / s.