Coefficient of friction question

[wtfcat]

Homework Statement

If a car's wheels are locked during emergency braking, the car slides along the road. The record for the longest skid marks on a road was reportedly set in 1960. the marks were 290m long. assuming that \mu = 0.6, how fast was the car going when the wheels became locked?

Givens:
V_f = 0m/s
V_i = ?
d = 290m
a = ?
\mu = 0.6
g = 9.8m/s²

Formulas:
\SigmaF = ma
\mu_k = F_k/F_n
V_f² = V_i² + 2ad

Homework Equations

1. Am i missing anything? Givens, formulas, etc.
2. Are my steps correct? ex. in my steps, am i missing a positive or negative sign, did i get units right, etc
3. Is my answer correct?

The Attempt at a Solution

So in this case, \SigmaF = F_k, correct? And F_k = \mu_kmg, so ma = \mu_kmg, and cancelling out the m gives me a = \mu_kg. so a = (0.6)(9.8) = 5.88m/s².
i sub "a" into the kinematic equation: V_f² = V_i² + 2ad ---> 0 = V_i² + 2(5.88)(290) ---> -V_i² = 3410.4 ---> -V_i = 58.3986m/s

If it's negative V_i, then does that mean that it was going backwards or something, or am i missing something here? Thanks in advance for taking the time to help me out!

 

[flyers]

No, you did everything right except that your car is decelerating so your a value should really be -5.88 and that would give you a positive vi value.

 

[wtfcat]

oh, ok, got it =)
thank you!