AboutAlon Mandes Expertise Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems.
Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
Experience 1. I'm a team member of mathnerds (math site for answering questions)
2. I'm a team member in the Student's Union of the Technion, helping
students who have problems in mathematics.
3. 2 years of experience as a math teacher in college.
4. I give free homework help for high school students in
Mathematics & Physics.
5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
"Complex Functions".
Question I am trying to investigate stopping distances as related to tailgating. Beginning drivers are often told to keep at least one car length behind the car in front of them for every 10 miles per hour of their speed. For example, a car traveling at 60 miles per hour should stay at least 6 car lengths away from the car in front. Is this true? Would a car traveling twice as fast need only keep twice as far away? I need to find out if a car traveling twice as fast need only keep twice as far back. For my model, I will assume that car brakes work by friction. As such, the velocity of the car, v, obeys the differential equation: v' = -kv, where k is a positive constant, called the coefficient of friction. Both cars have the same coefficient of friction. I am assuming also that the time it takes me to notice that the car in front is braking and to apply my brakes, b, is constant and doesn't depend of my speed. If both cars are initially travelling at speed, s, how much distance should I leave between me and the car in front to be sure that I can stop without hitting them. My answer must be a function d of k, s, and b. Does d(k, 2s, b) = 2d(k, s, b)?
Answer Certainly not ! d(2v) Not Equal 2d(v). In order to understand the
relationship between the distance & the velocity, we relay on the
law of Energy conservation : Kinetic Energy = Frictional energy(heat)
when breaking. So, 0.5mv^2=umgx. Where g=9.8m/sec^2, m the mass
of the car, v is the speed at breaking, u is the friction coefficient & x is the distance till full stop. From this equation we obtain the relation x=(constant)*v^2. meaning , if v is multiplied by 2 then the distance grows by 4 !!
