Verification of a problem -> force needed for car stopping in 1.8cm

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In summary: Then, you have three practical limitations to that strong braking effect:1) Friction between tires and asphalt.2) Excessive negative acceleration (g-force) for passengers.3) Excessive heat in brake pads-discs.
  • #1
simphys
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Homework Statement
picture
Relevant Equations
newton's 2nd law of motion F_net = ma
Kinematic equation: v_fx^2 = v_0x^2 + 2*a_x*(deltaX)
1644778250216.png

hello guys can someone verify my solution (in the picture) whether the solution is correct or I guess the steps followed would be enough. There is no solution in the book for this one but wanted to be sure.

Thanks in advance
 

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  • #2
You used 37 m/s for the initial velocity of the car when it is given as 37 km/h. The method is correct.
 
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  • #3
Why does your homework statement talk of a 850 kg car traveling at 45 km/h when the question you copy says a 840 kg car traveling at 37 km/h?
Note v0 = 37 km/h, NOT 37 m/s!
First rule of any test is READ THE QUESTION ACCURATELY.

Edit; kuruman got there first!
 
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  • #4
kuruman said:
You used 37 m/s for the initial velocity of the car when it is given as 37 km/h. The method is correct.
mjc123 said:
Why does your homework statement talk of a 850 kg car traveling at 45 km/h when the question you copy says a 840 kg car traveling at 37 km/h?
Note v0 = 37 km/h, NOT 37 m/s!
First rule of any test is READ THE QUESTION ACCURATELY.

Edit; kuruman got there first!
Thanks guys, appreciate it! And well no.. I mistakenly copied the wrong question. I typed it in google but found a question on cheg from the same book (other edition) with just the numbers changed hence why, my apologies
 
  • #5
Then, you have three practical limitations to that strong braking effect:
1) Friction between tires and asphalt.
2) Excessive negative acceleration (g-force) for passengers.
3) Excessive heat in brake pads-discs.
 
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  • #6
Lnewqban: That's why you need to supply your own brick wall (or borrow a neighbor's). The car *can* stop on a dime, though.
 
  • #7
I don't see why advertise a car that can stop on a dime. Its air bags would have to be deployed every time it did so. If it doesn't have air bags for that reason, its driver can stop on a dime exactly once in a lifetime.
 
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  • #8
Lnewqban said:
Then, you have three practical limitations to that strong braking effect:
1) Friction between tires and asphalt.
2) Excessive negative acceleration (g-force) for passengers.
3) Excessive heat in brake pads-discs.
Thanks for the indication that.
Mike S. said:
Lnewqban: That's why you need to supply your own brick wall (or borrow a neighbor's). The car *can* stop on a dime, though.
I presume that that was sarcastic😄
 
  • #9
simphys said:
Thanks for the indication that.
Velocity used in the calculation should be 10.28 m/s.

After you calculate the force necessary to stop in 1.8 cm, just for curiosity, you could see how much force the maximum available static friction (rubber-asphalt μ is around 0.8) would actually allow, as well as how big of g-force the passengers would experience.
 
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  • #10
kuruman said:
its driver can stop on a dime exactly once in a lifetime.
While this may be true for the spirit of the idiom, if taken literally the driver can do it many many times. Stopping on a dime would technically just require you to be on top of it as you come to a rest. 😏
 
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  • #11
Lnewqban said:
Velocity used in the calculation should be 10.28 m/s.

After you calculate the force necessary to stop in 1.8 cm, just for curiosity, you could see how much force the maximum available static friction (rubber-asphalt μ is around 0.8) would actually allow, as well as how big of g-force the passengers would experience.
wow that's insane very curious how this these coefficients for the frictions are calculated (I guess looking at the molecular interaction between both surfaces explains that.. and lots of experimentation ofc)
 
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FAQ: Verification of a problem -> force needed for car stopping in 1.8cm

What is the purpose of verifying a problem related to the force needed for a car to stop in 1.8cm?

The purpose of verifying a problem related to the force needed for a car to stop in 1.8cm is to ensure that the calculations and assumptions being made are accurate and reliable. This is important for ensuring the safety and efficiency of the car's braking system.

How is the force needed for a car to stop in 1.8cm calculated?

The force needed for a car to stop in 1.8cm is calculated using the formula F = m x a, where F is the force, m is the mass of the car, and a is the deceleration. This calculation takes into account the weight and speed of the car, as well as the friction between the tires and the road.

What factors can affect the force needed for a car to stop in 1.8cm?

Several factors can affect the force needed for a car to stop in 1.8cm, including the weight and speed of the car, the condition of the tires and brakes, the road surface and weather conditions, and the presence of any obstacles or hazards on the road.

Why is it important to consider the force needed for a car to stop in 1.8cm?

It is important to consider the force needed for a car to stop in 1.8cm because it directly affects the safety of the driver, passengers, and other individuals on the road. If the force is too high, it could result in a collision or loss of control of the vehicle. If the force is too low, the car may not be able to stop in time to avoid an accident.

How can the force needed for a car to stop in 1.8cm be reduced?

The force needed for a car to stop in 1.8cm can be reduced by maintaining the car's braking system and tires in good condition, driving at a safe speed, and being aware of potential hazards on the road. Additionally, using advanced braking technologies such as anti-lock brakes can also help reduce the force needed to stop a car in a short distance.

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