2 second rule true at highway speeds?

  • Thread starter Thread starter rarouch
  • Start date Start date
AI Thread Summary
The two-second rule for following distance may not be sufficient at highway speeds due to the time it takes for a driver to react and brake. At 100 km/h, a vehicle traveling two seconds behind would cover approximately 60 meters, but a typical reaction time of 0.5 seconds adds an additional 15 meters of travel before braking begins. This means that the following car may still be too close to avoid a collision if the lead car brakes suddenly. The braking rates of both vehicles are crucial; if they are similar, the two-second rule may be adequate, but differences in braking efficiency could lead to danger. Overall, maintaining a greater distance than two seconds is advisable to account for reaction time and braking dynamics.
rarouch
Messages
5
Reaction score
0
Here's the question:

A safe following distance for a car is often described as "two seconds". In
other words, if you pass the same point as the car in front of you two seconds
later, the distance between the cars is safe.
Inquiry: is this rule true at highway speeds?

Parameters:

Car A is in front, car B is behind.
Both are traveling with the same velocity (approx 100 km/h).
There are two seconds between car A and car B.
Car A will commence braking at a rate you will be given later.
Car B will not react for a certain amount of time (given later) and then will
brake.
The rate of car B's braking will be given later.

Question: Can car B stop before it collides with car A?

I was given the question this way. It's basically an evaluation that my teacher will give on a day and I have time to prepare for it. What would be reasonable amounts for the parameters?

My thoughts:
It basically depends on the rate of the braking of the cars. But in general, the car's deceleration will be approximately same and since we can take that the car B driver will react quickly than 2 sec therefore they may not collide but they really depend on the rates, which is why I was wondering, what would be a reasonable breaking rate? and how would I unput that into a formula?

Thank you very much.
 
Physics news on Phys.org
rarouch said:
Here's the question:

A safe following distance for a car is often described as "two seconds". In
other words, if you pass the same point as the car in front of you two seconds
later, the distance between the cars is safe.
Inquiry: is this rule true at highway speeds?

Parameters:

Car A is in front, car B is behind.
Both are traveling with the same velocity (approx 100 km/h).
There are two seconds between car A and car B.
Car A will commence braking at a rate you will be given later.
Car B will not react for a certain amount of time (given later) and then will
brake.
The rate of car B's braking will be given later.

Question: Can car B stop before it collides with car A?

I was given the question this way. It's basically an evaluation that my teacher will give on a day and I have time to prepare for it. What would be reasonable amounts for the parameters?

My thoughts:
It basically depends on the rate of the braking of the cars. But in general, the car's deceleration will be approximately same and since we can take that the car B driver will react quickly than 2 sec therefore they may not collide but they really depend on the rates, which is why I was wondering, what would be a reasonable breaking rate? and how would I unput that into a formula?

Thank you very much.
Once you know the braking rate (deceleration) of each car, you'll have to use the kinematic motion equations. A car can only decelerate as fast as friction allows; and friction depends on the tire type and road surface conditions. Oh, maybe 6 m/s/s might be an approximate rate on dry pavement, without the tires skidding.
 
PhanthomJay said:
Once you know the braking rate (deceleration) of each car, you'll have to use the kinematic motion equations. A car can only decelerate as fast as friction allows; and friction depends on the tire type and road surface conditions. Oh, maybe 6 m/s/s might be an approximate rate on dry pavement, without the tires skidding.

Thank you for the input. What would I have to do if the braking rate for both of the cars are different?
 
Originally you might assume both cars slow due to braking at the same deceleration.

The hint you need is this:

"Car B will not react for a certain amount of time (given later) and then will brake".

For car B, assume reaction time is constant, independent of velocity, and initially assume 2 seconds reaction time. How is the following-distance to avoid collision a function of the velocities of car A and B?
 
Last edited:
rarouch said:
Here's the question:

A safe following distance for a car is often described as "two seconds". In
other words, if you pass the same point as the car in front of you two seconds
later, the distance between the cars is safe.
Inquiry: is this rule true at highway speeds?

Parameters:

Car A is in front, car B is behind.
Both are traveling with the same velocity (approx 100 km/h).
There are two seconds between car A and car B.
Car A will commence braking at a rate you will be given later.
Car B will not react for a certain amount of time (given later) and then will
brake.
The rate of car B's braking will be given later.

Question: Can car B stop before it collides with car A?

I was given the question this way. It's basically an evaluation that my teacher will give on a day and I have time to prepare for it. What would be reasonable amounts for the parameters?

My thoughts:
It basically depends on the rate of the braking of the cars. But in general, the car's deceleration will be approximately same and since we can take that the car B driver will react quickly than 2 sec therefore they may not collide but they really depend on the rates, which is why I was wondering, what would be a reasonable breaking rate? and how would I unput that into a formula?

Thank you very much.

Good brakes and a dry road would generally give a braking rate of 0.5g or -5 m.s-2.

Given that 100 km/h is just under 30 m.s-1, that means nearly 6 seconds to stop. That means a braking distance of nearly 90m - but both cars take that long to brake so that is not the problem.
It is the reaction distance - the distance covered by the second car while the driver reacts.

If you are 2 seconds behind the car in front, you would be almost 60m behind at highway speed.
A typical reaction time is just under 0.5 seconds.
In that time you will travel ~ 15m. so you should still stop 45 m from the car in front.

If the following car has less efficient brakes, it will be a bit closer.

Note: if the braking efficiency is the same for both cars, the distance behind the car in front needs to be your reaction time at least, so traveling 2 seconds behind allows for your reaction time to be a hopeless 2 seconds.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top