Kinematics - Constant acceleration and friction

In summary, the problem presents a car traveling at a constant speed on a flat highway and asks for the minimum distance needed for the car to stop, given the coefficient of kinetic friction. However, with missing variables such as time and mass, the problem seems unsolvable. After receiving guidance and realizing that there is no opposing force to friction, the problem is solved by calculating the acceleration using Newton's equation of motion and then using it to find the distance traveled using the equation for displacement. The mass of the car is also determined to be necessary for solving the problem.
  • #1
Kaoi
21
0
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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  • #2
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.
 
  • #3
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...
 
  • #4
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 [tex]v_{final}=v_{0}-at[/tex] 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, [tex]x(t)=v_{0}t-\frac{1}{2}at^2[/tex], to find the distance which the car passes until it stops.
 
  • #5
If you don't know the mass of the car, the "coefficient of friction" is irrelevant.
 
  • #6
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.
 
  • #7
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!
 

1. What is constant acceleration in kinematics?

Constant acceleration refers to the rate at which an object's velocity changes over time, which remains the same throughout the object's motion. This means that the object's speed increases or decreases by the same amount over each unit of time.

2. How is constant acceleration calculated?

Constant acceleration can be calculated using the formula a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval. This formula assumes that the acceleration remains constant throughout the object's motion.

3. What is the significance of friction in kinematics?

Friction is a force that opposes the motion of an object and acts in the opposite direction of the object's motion. In kinematics, friction is important because it affects the acceleration and velocity of an object. It can also cause an object to come to a stop or change direction.

4. How does friction affect an object's motion?

Friction can either increase or decrease an object's acceleration, depending on whether it acts in the same direction or opposite direction as the applied force. It can also cause an object to slow down or stop due to the opposing force it creates. Additionally, friction can cause an object to change direction if it is acting at an angle to the applied force.

5. What are some real-life examples of constant acceleration and friction?

A common example of constant acceleration and friction is a car accelerating on a road. The car's acceleration is constant as it gains speed, but friction from the road's surface and air resistance slows the car down. Another example is a ball rolling down a ramp, where gravity provides constant acceleration, but friction from the ramp's surface slows the ball down. Friction is also present in activities such as walking, biking, and ice skating.

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