Determine the maximum theoretical speed expected for a car

[Apprentice123]

Determine the maximum theoretical speed expected for a car, leaving the rest, covering a distance of 50m. The coefficient of static friction between tire and road is 0,80. Knowing that the front wheels bear 60% of the weight of the car and the back, the remaining 40%. Determine the speed (a) traction front (b)
traction back


Answer:
(a) 78,1 Km/h
(b) 63,8 Km/h

 

[LowlyPion]

The answers look OK.

 

[Apprentice123]

The answers look OK.

The answers are in the book. I do not understand the problem

 

[cepheid]

If a car is to reach a final speed v_f over a distance of d = 50 m at constant acceleration, then you know from basic kinematics what acceleration, a, is required.

However, the acceleration is the driving force over the mass. In this case, the driving force is equal to the frictional force between the tires and the road (this is what propels the vehicle forward). If the acceleration to reach v_f is too high, the tires will not be able to provide enough static friction to support this acceleration, and the tires will slip (we say that the car loses traction).

Therefore, the question is asking what is the maximum allowable v_f (at the end of 50 m) that will ensure that the driving tires maintain traction? The answer is different for the front tires and the rear tires, because each set of tires is bearing a different load and is therefore able to provide a different amount of static friction force.

 

[Apprentice123]

If a car is to reach a final speed v_f over a distance of d = 50 m at constant acceleration, then you know from basic kinematics what acceleration, a, is required.

However, the acceleration is the driving force over the mass. In this case, the driving force is equal to the frictional force between the tires and the road (this is what propels the vehicle forward). If the acceleration to reach v_f is too high, the tires will not be able to provide enough static friction to support this acceleration, and the tires will slip (we say that the car loses traction).

Therefore, the question is asking what is the maximum allowable v_f (at the end of 50 m) that will ensure that the driving tires maintain traction? The answer is different for the front tires and the rear tires, because each set of tires is bearing a different load and is therefore able to provide a different amount of static friction force.

What is driving force ?
V^2 = Vo^2 + 2a(x - xo)

V = ?
Vo = 0
x - xo = 50
a = How to calculate?

 

[LowlyPion]

What is driving force ?
V^2 = Vo^2 + 2a(x - xo)

V = ?
Vo = 0
x - xo = 50
a = How to calculate?

Consider the friction available through the wheels, to provide motive force for the vehicle. How much of force, given the weight distribution between the wheels and the coefficient of friction through say the back wheels, can the back wheels provide to move the vehicle forward? Keep in mind that if the engine supplies too much power that a great deal of the power will go to burning rubber. So there is an upper limit that you can determine through the weight and friction relationship.

What they want is how fast can it accelerate if power comes from the rear wheels, and then how much if supplied through the front.

 

[Apprentice123]

Consider the friction available through the wheels, to provide motive force for the vehicle. How much of force, given the weight distribution between the wheels and the coefficient of friction through say the back wheels, can the back wheels provide to move the vehicle forward? Keep in mind that if the engine supplies too much power that a great deal of the power will go to burning rubber. So there is an upper limit that you can determine through the weight and friction relationship.

What they want is how fast can it accelerate if power comes from the rear wheels, and then how much if supplied through the front.

Yes But I do not know which formula I use

 

[cepheid]

Yes But I do not know which formula I use

Two of us have told you that the force propelling the car forward is equal to the FRICTION force between the tires and the road. What is the formula for a friction force?

 

[Apprentice123]

Not find the answer:

Sum of forces (Z)
* u = 0,80

ZFy = 0
N - P = 0
N = P


ZFx = m.a
-N.u + V = m.a
-(m.g.u) + V = m.a
V = a + 7,848


V^2 = (Vo)^2 + 2.a.x
...
a = 84,3 m/s^2

V = 92,148 m/s^2

60% V => 55,288 m/s
40% V => 36,86 m/s

 

[cepheid]

ZFx = m.a
-N.u + V = m.a
-(m.g.u) + V = m.a
V = a + 7,848

No, you cannot have velocity in the force balance equation. Velocity is not a force!

Try this. Newton's Third Law says that if the tires push back on the road, then the road pushes forward on the tires. So the forward force on the car, F_forward is:

F_forward = μN

ΣF_x = F_forward = μN = ma

Now you can calculate a.

Be careful! N is not the same for the front wheels as it is for the back wheels.

 

[Apprentice123]

No, you cannot have velocity in the force balance equation. Velocity is not a force!

Try this. Newton's Third Law says that if the tires push back on the road, then the road pushes forward on the tires. So the forward force on the car, F_forward is:

F_forward = μN

ΣF_x = F_forward = μN = ma

Now you can calculate a.

Be careful! N is not the same for the front wheels as it is for the back wheels.

Thank you very much

A) (m.g.u.60)/100 = m.a
a = 4,7088 m/s^2

V^2 = (Vo)^2 + 2aX
V = 21,7 m/s => 78,12 Km/h

B) (m.g.u.40)/100 = m.a
a = 3,1392 m/s^2
V = 17,71 m/s => 63,756 Km/h