Kinematics - Constant acceleration and friction

AI Thread Summary
The discussion revolves around calculating the minimum stopping distance for a car traveling at 52.4 km/hr on a flat highway with a coefficient of kinetic friction of 0.151. Initially, the poster felt the problem was unsolvable due to missing variables, particularly time and mass. However, it was clarified that the deceleration can be determined using the coefficient of friction and gravitational force, allowing for the calculation of stopping distance. The final solution involved using the kinematic equation vf² = vi² + 2aΔx to find the distance once the correct acceleration was established. Ultimately, the problem was resolved after recognizing the relationship between friction and deceleration.
Kaoi
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Unsolvable? Kinematics - Constant deceleration and friction

Here's the way the problem is laid out:

A car travels at 52.4 km/hr on a flat highway. If the coefficient of kinetic friction is 0.151, what is the minimum distance needed for the car to stop?

Given:
Vi = 52.4 km/hr (calculated as 14.555(...) m/s)
Vf = 0 m/s
muk= 0.151
g = 9.81 m/s2

Unknown:
/\ x, /\ t, a, m

It seems to me that I have too many missing variables to solve the problem.

I know that Vavg = (Vi + Vf)/2, so the average velocity would be around 7. 277 m/s, but without /\t, I can't carry out any of the normal kinematic equations to find /\x, and without m, I can't figure out Fk, Fg, or FN.

Is there something I'm missing, or is the problem actually unsolvable?
 
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Why don't you think you can find t? You know the average speed the car will have over the distance. If you have the average rate of change of distance with respect to time, and you know the total distance then you can determine time.
 
But that's just the problem-- I'm trying to find the distance (/\x), and to do that, I would need a, which requires /\t, none of which I have...
 
Kaoi said:
Here's the way the problem is laid out:

A car travels at 52.4 km/hr on a flat highway. If the coefficient of kinetic friction is 0.151, what is the minimum distance needed for the car to stop?

Given:
Vi = 52.4 km/hr (calculated as 14.555(...) m/s)
Vf = 0 m/s
muk= 0.151
g = 9.81 m/s2

Unknown:
/\ x, /\ t, a, m

It seems to me that I have too many missing variables to solve the problem.

I know that Vavg = (Vi + Vf)/2, so the average velocity would be around 7. 277 m/s, but without /\t, I can't carry out any of the normal kinematic equations to find /\x, and without m, I can't figure out Fk, Fg, or FN.

Is there something I'm missing, or is the problem actually unsolvable?

You can calculate the decceleration of the car from Newton's equation of motion, where the decceleration must be proportional to the resultant force acting on the car, which is the force of friction. Generally, this force equals F = k N, where k is the coefficient of friction (dynamic friction, in your case), and N is the reaction from the ground acting on the car, which equals the weight of the car. Now, since you know the decceleration, you can use the equation v_{final}=v_{0}-at to calculate the time of decceleration from 52.4 km/hr to 0 km/hr. In the end, you can use the equation for displacement of the car, x(t)=v_{0}t-\frac{1}{2}at^2, to find the distance which the car passes until it stops.
 
If you don't know the mass of the car, the "coefficient of friction" is irrelevant.
 
HallsofIvy said:
If you don't know the mass of the car, the "coefficient of friction" is irrelevant.

Absolutely right, I oversaw that the mass is not given.
 
Solved!

After asking my teacher, I solved the problem-- I was assuming that there was a force in the opposite direction of friction.

Basically, the acceleration was

-(mu(k)Fn)/m, which equals

-(mu(k)mg)/m, canceling to

a = -(mu(k)g).

Then I could use the vf2 = vi2 + 2a/\x equation to solve for distance.

Thank you all for your help anyway!
 
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