How to calculate the distance a car skids before coming to a halt?

In summary, the driver of a car travelling at a speed of 58.0 km/h suddenly locks the wheels by slamming on the brakes. With a coefficient of kinetic friction of 0.720 between the tires and the road, the car skids a distance of 0.5 * (58.0 km/h)^2 / (0.720 * 9.8 m/s^2) = 106.5 meters before coming to a halt. To solve this problem, you can use Newton's second law and kinematic equations to determine the acceleration and time of deceleration, and then use the formula for distance travelled with uniform acceleration. The mass of the car is not needed to find the distance since it cancels
  • #1
frozen7
163
0
Travelling at a speed of 58.0 km/h, the driver of an automobile suddenly locks the wheels by a slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.720. How far does the car skid before coming to a halt? Ignore effects of air resistence.

How to solve this problem? I have no idea at all how to start since seem like lack of some others value. We have only the value of v and coefficient of kinetic friction. How to solve this??
 
Physics news on Phys.org
  • #2
Start by drawing a FreeBody Diagram (find the acceleration, using Newton's 2nd Law), and then work the time with the uniform acceleration equations from kinematics.
 
  • #3
How to find out acceleration as we don`t know the mass of the car?
 
  • #4
You don't need the mass of the car, why don't you set up the equations algebraicly and see where it takes you?
 
  • #5
Or you can use the conservation of energy principle.
 
  • #6
Be v=58 km/h the velocity and f=0.720 the friction coefficient.

The only force acting on the car is friction. Newton tells us that the sum of all forces acting on a point of mass equals m*a. So you know the friction equals m*a. That is (on a horizontal plane):

[tex]m\cdot a=f\cdot m \cdot g[/tex]
Divide by m on both sides and you get:
[tex]a=f\cdot g[/tex]

Now you know the acceleration.

Be the time at the moment, where the car starts breaking (the wheels stand still), [tex]t_{i}=0[/tex] and the distance at that point of time [tex]s_{i}=0[/tex].
Notice: we have reduced the problem thru a dynamic consideration (fricition is the only force) to a purely kinematic one. We have now an accelerated motion (no initial distance and no movement with a constant speed). That's it. So we use the kinematic formulas you're surely familiar with:

[tex]s=\frac{1}{2}\cdot a t^2[/tex]
We also know that acceleration a=v/t. Therefore, t=v/a. We can thus compute s:
[tex]s=\frac{1}{2}\cdot a \left(\frac{v}{a}\right)^2=\frac{1}{2}\cdot \frac{v^2}{a}[/tex]

That was it.
 
  • #7
Thanks..but I am still not really understand why ma = f (mg). Even the car is moving with a deceleration, isn`t there any force which cause the car moving forward?
Some more, why consider the initial speed = 0 when finding the distance?
 
  • #8
frozen7 said:
Thanks..but I am still not really understand why ma = f (mg).
The only horizontal force on the car is that of friction. And the value of the kinetic friction force is [itex]F_f = \mu N[/itex], where N is the normal force between the surfaces (the tires and the road). In this case, the normal force is just the weight of the car (mg). Note that the mass cancels, so you don't need to know what it is.
Even the car is moving with a deceleration, isn`t there any force which cause the car moving forward?
A force is only needed to change the motion. If there were no forces acting to slow the car down, the car would keep moving forward.
Some more, why consider the initial speed = 0 when finding the distance?
It may be less confusing if you took the initial speed to be 58.0 km/h, the final speed to be 0, and the acceleration to be negative. (Don't forget to change everything to standard units. For example, speed should be in m/s.)
 

1. What is the "Physics car breaking problem"?

The "Physics car breaking problem" is a hypothetical scenario used in physics to illustrate the concept of inertia and how it affects the motion of objects. It involves a car traveling at a constant speed and suddenly braking to a stop.

2. What is inertia and how does it relate to the "Physics car breaking problem"?

Inertia is the tendency of an object to resist changes in its state of motion. In the "Physics car breaking problem", the car's inertia causes it to continue moving forward even after the brakes are applied, resulting in a longer stopping distance.

3. What factors affect the stopping distance of a car in the "Physics car breaking problem"?

The stopping distance of a car in the "Physics car breaking problem" is affected by the speed of the car, the mass of the car, and the friction between the car's tires and the road surface. The higher the speed and mass of the car, and the lower the friction, the longer the stopping distance will be.

4. Can the stopping distance of a car be reduced in the "Physics car breaking problem"?

Yes, the stopping distance of a car can be reduced by increasing the friction between the car's tires and the road surface. This can be achieved by using tires with better traction or by improving the road surface.

5. How does the "Physics car breaking problem" apply to real-life situations?

The "Physics car breaking problem" is a simplified example of the physics involved in a car's braking system. In real-life situations, other factors such as reaction time, road conditions, and the condition of the car's brakes also play a role in determining the stopping distance of a car. However, the concept of inertia and its effects on stopping distance still apply.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
717
  • Introductory Physics Homework Help
Replies
8
Views
5K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
3K
Back
Top