Solving for time and displacement

[Tabe]

Ok, here's the problem that I am working on now.
Sarah is walking at a speed of 1.5m/s when she accidentally slips on a patch of ice whose coefficient of kinetic friction is 0.060. If Sarah has a mass of 40.0 kg, how long before she comes to a stop on the ice?
I'm stuck because I don't know which equation to use to solve for time or displacement. Each equation that I have has both variables in it. Also, where does the coefficient of kinetic friction factor into the equation. I solved for the normal force, and force of friction, but I still ended up in a dead end.

 

[HallsofIvy]

Yes, the equations you have:

F= ma: Force= mass*acceleration. You know Sarah's mass so you can calculate Sarah's weight so you can calculate the Force slowing Sarah down. From F= ma, you can calculate the acceleration. Apparently you have already done that: great!

v(t)= at+ v₀
velocity at time t, with initial velocity v₀= 1.5 m/s.

x(t)= (a/2)t²+ v₀t
x(t) is the distance Sarah went in time t.

You're right: the second equation involves both x and t. But the first equation doesn't involve x. You know that Sarah's final speed is 0 ("she comes to a stop") so at+ 1.5= 0. You know a, so you can find t.

Now that you know t, use the second equation to calculate x.

 

[wais]

To solve for time and displacement in this problem, you will need to use the equation for Newton's Second Law of Motion: F = ma. In this case, the force acting on Sarah is the force of friction, which is equal to the coefficient of kinetic friction (μ) multiplied by the normal force (mg). So the equation becomes F = μmg.

To solve for time, you will need to use the equation for acceleration: a = F/m. Plug in the values for the force of friction and Sarah's mass, and you will get the acceleration (a) on the ice.

Next, you can use the equation for displacement: d = v0t + 1/2at^2. In this case, v0 is Sarah's initial velocity, which is 1.5m/s. You can set the final displacement (d) to be 0, since Sarah will come to a stop on the ice. You can also plug in the acceleration (a) that you calculated earlier. This will give you an equation with only one variable, time (t).

To solve for time, you can use the quadratic formula: t = (-b ± √(b^2-4ac))/2a. In this case, a = 1/2a, b = v0, and c = -d. Plug in the values and solve for t. This will give you the time it takes for Sarah to come to a stop on the ice.

To solve for displacement, you can use the same equation, but plug in the value you got for time (t) and solve for d. This will give you the distance Sarah traveled before coming to a stop on the ice.

Remember to always check your units and make sure they are consistent. Also, don't forget to include the units when writing your final answer. I hope this helps you solve the problem!