a) The hockey puck's velocity function v(t) can be found by considering Newton's second law, F = ma, where F is the net force acting on the puck and a is its acceleration. In this case, the net force is the friction force, F_r = -mu*m*g. Since the acceleration a is the derivative of velocity with respect to time, we have:
ma = m(dv/dt) = -mu*m*g
Dividing both sides by m, we get:
dv/dt = -mu*g
To find the velocity function v(t), we need to solve this differential equation. Rearranging terms, we have:
dv = -mu*g*dt
Integrating both sides, we get:
∫dv = ∫-mu*g*dt
v(t) = -mu*g*t + C
To find the constant C, we use the initial condition v(0) = v_0:
v_0 = -mu*g*0 + C
C = v_0
So the velocity function is:
v(t) = -mu*g*t + v_0
b) To find the time t_rest when the puck comes to rest, we set v(t) = 0:
0 = -mu*g*t_rest + v_0
Solving for t_rest, we get:
t_rest = v_0 / (mu*g)
c) The position function x(t) can be found by integrating the velocity function v(t) with respect to time:
x(t) = ∫v(t) dt = ∫(-mu*g*t + v_0) dt
x(t) = -mu*g*(t^2)/2 + v_0*t + D
To find the constant D, we use the initial condition x(0) = 0:
0 = -mu*g*(0^2)/2 + v_0*0 + D
D = 0
So the position function is:
x(t) = -mu*g*(t^2)/2 + v_0*t
d) To find the distance travelled x_rest before the puck comes to rest, we need to find the value of x(t) at t = t_rest:
x_rest = x(t_rest) = -mu*g*(t_rest^2)/2 + v_0*t_rest
Substituting the expression for t_rest from part (b), we get:
x_rest = -mu*g*((v_0 / (mu*g))^2)/2 + v_0*(v_0 / (mu*g))
x_rest = (v_0^2) / (2*mu*g)
So the distance travelled by the hockey puck before coming to rest is:
x_rest = (v_0^2) / (2*mu*g)