Exploring Engineering Chapter 13 Engineering Kinematics

Kinematics and Traffic Flow

Transportation Transportation Transportation Engineering The movement of goods and people Transportation Engineering Design and operation of facilities and vehicles to enhance transportation

Topics Covered Speed and acceleration Kinematics: the relationships between distance, velocity/ speed, acceleration,and time Highway capacity

Speed and Acceleration Speed is not the same thing as “velocity.” Both have dimensions of distance/time but “speed” is only the magnitude of “velocity” In one-dimensional problems, they are equivalent In multidimensional problems, “velocity” connotes direction as well as “speed”; here we will only deal in one-dimensional problems The average speed is calculated as Δx/Δt, where Δx is the distance traveled in time Δt. Acceleration has dimensions of speed/time, and the average acceleration can be calculated as ΔV/Δt. It is the “slope” of a speed vs. time graph. To calculate the instantaneous speed or acceleration, the “Δ”s used in these equations should be very, very small. Speed units: US (miles per hour) Rest of the world (kilometers per hour) SI units: meters per second (m/s)

Example A student sees his/her bus coming down the street and starts running at 2.5 m/s toward the bus stop. The bus is traveling at 10.0 m/s. The student starts running toward the bus stop when the bus is 50. m behind. What is the maximum distance the student can be from the bus stop to catch the bus? Know: A) Student runs at 2.5 m/s. B) Bus is traveling at 10.0 m/sc. C) Distance between them 50. m Find: How far can the student be from the bus stop?

Solution How: In time Δt the bus travels a distance Δxb= Vb  Δt, and the student travels a distance Δxs= Vb  Δt. If they both reach the bus stop at the same time, then the bus will have traveled a distance 50. meters more than the student travels, or Δxb= Δxs+ 50 .m Since Δxb= Vb  Δt and Vb  Δt then Vb  Δt = Vb  Δt + 50. m, or (Vb - Vs)(Δt) = 50. m Solve: Solving for Δt we get Δt = 50./(Vb - Vs) = 50./(10. – 2.5) = 6.7 seconds

Distance vs. Time chart The slope of the line is the speed.

Average Speed Average speed is a measure of the distance traveled in a given period of time. Suppose that during your trip to school, you traveled a distance of 5.0 miles and the trip lasted 0.20 hours (12. minutes). The average speed of your car could be determined as On the average, your car was moving with a speed of 25. miles per hour. During your trip, there may have been times that you were stopped and other times that your speedometer was reading 50. miles per hour; yet on the average you were moving with a speed of 25. miles per hour.

Kinematics of Motion “Equations of motion” Simplified case: constant acceleration x = Vot+ (1/2)at2 (a = acceleration) V= dx/dt (or Δx/ Δt, slope of distance vs. time curve) So, V= Vo + at a = dV/dt (or ΔV/Δt, slope of velocity vs time curve) So, a = (dV/dx)(dx/dt) = (dV/dx)V We will not be using these physics results

Graphical View The various regions in this graph represent Stationary, Constant speed, and Variable speed (i.e., acceleration)

Constant Velocity Algebraic Method Apply Cartesian geometry Note: speed is distance/time Dimensions are length/time Typical units ft/s, m/s Origin Distance Time x0 t0 t x

Constant Velocity Calculus Method Speed is the rate of change of distance with time We can integrate from the initial conditions (x0, t0) since V is constant

Graphical Method Consider the various regions in this graph Variable Origin Speed Time Constant speed, V0 acceleration Variable Consider the various regions in this graph Constant speed Constant acceleration Variable acceleration

Constant Acceleration Algebraic Method Origin Speed Time Constant speed, V0 t0 t V Vaver

Use of speed/time diagram We have already seen that the slope of the speed vs. time is acceleration The area under the speed/time curve between t and to is the distance covered in that time interval Origin Speed Time Constant speed, V0 t0 t V Vaver

Example You are designing an automated highway using vehicle speeds of 100. mph. How long does the on-ramp need to be to allow the car to reach this speed and how long will it take the vehicle to accelerate to this speed? Assume the vehicle will start at V0 = 0 and a = 5.00 ft/s2 at t = 0 sec.

Example Continued a = 5.00 ft/s Speed, ft/s t sec Need: x at V = 100. mph and t at 100 mph. Note: 100 mph = 5280 ×100./3600. [ft/mile] [miles/hr] [hr/s] = 147 ft/s Know: a = 5.00 ft/s2, V0= 0 ft/s at t0 = 0 s How? Use a speed time graph 1) V/t = a or t = 147/5.00 = 29.4 s gives the time 2) Area of triangle = ½  147  29.4 [ft/s] [s] = 2160 ft  gives distance 147 ft/s a = 5.00 ft/s Speed, ft/s t sec

Sneaky Example Suppose you want to travel a distance of 2.0 miles at an average speed of 30. mph. You cover the first mile at a speed of 15. mph. What should your speed be in the second mile so you will average 30. miles per hour for the entire trip? (Hint: It’s faster than you think!)

Sneaky Example Continued 30. Speed, mph 15. Speed, mph 2.00 mile 1.00 mile t, s t1 t, s t2 Area = 1.00 miles = 15. t0 or t0 = 1/15. miles = 4.0 minutes Area = 2.00 miles = 30.  t0 or t0 = 2.00/30. miles = 4.0 minutes

Sneaky Example Moral So the second mile will have to be covered in zero s! Or infinite speed! The lesson is you can’t average averages! Only time and distance are preserved quantities!

Subway Example acceleration = 6.0 ft/s2, deceleration = - 4.0 ft/s2, A subway train is being planned using trains capable of 50.0 mph. How close can adjacent stations be so that the train will reach a speed of 50.0 mph given these characteristics: acceleration = 6.0 ft/s2, deceleration = - 4.0 ft/s2, maximum speed is 50.0 mph, and the train starts at V0 = 0. As we talk about smart cars and zero emission vehicles it is important to realize that cars are not the best way to move people. Subways and mass transit systems can move large quantities of people efficiently. It is important to determine the types of equipment which will be needed for a subway system. The comfortable acceleration and deceleration rate for standing people is 4 - 6 fps^2. For seated people this is increased to 8 - 10 fps^2

Subway Example Need: x1 and x2 corresponding to t1 and t2 Know: V0 (0) = 0 and V(t2) = 0; a1 = 6.0 ft/s2 and d2 = - 4.0 ft/s2, V1(t1) = V2(0) = 50. mph = 73 ft/s How: Speed/time graph We know that x1 + x2 = 1320 ft

Subway Example 6.0 ft/s2 73. Speed, ft/s -4.0 ft/s2 t0= 0 t1 t2 As we talk about smart cars and zero emission vehicles it is important to realize that cars are not the best way to move people. Subways and mass transit systems can move large quantities of people efficiently. It is important to determine the types of equipment which will be needed for a subway system. The comfortable acceleration and deceleration rate for standing people is 4 - 6 fps^2. For seated people this is increased to 8 - 10 fps^2 t0= 0 t1 t2

Subway Example Solution: part 1:V1 = 73. ft/s t1 = 73. /6.0 [ft/s] × [s2/ft] = 12.2 s Area = x1 = ½ × 73 × 12.2 [ft/s][s] = 445 ft Similarly for t2 - t1 = -73./-4.0 [ft/s][s2s] = 18.3 s Area = x2 = ½ × 73 × 18.3 [ft/s][s] = 668 ft Station separation = 445 + 668 = 1,110 ft  We know that x1 + x2 = 1320 ft 

Highway Capacity - Types of Traffic Flow The first type is called uninterrupted flow, and is flow regulated by vehicle-vehicle interactions and interactions between vehicles and the roadway. For example, vehicles traveling on an interstate highway are participating in uninterrupted flow. The second type of traffic flow is called interrupted flow. Interrupted flow is flow regulated by an external means, such as a traffic signal. Under interrupted flow conditions, vehicle-vehicle interactions and vehicle-roadway interactions play a secondary role in defining the traffic flow. The material on this subject is from http://www.webs1.uidaho.edu/niatt_labmanual/Chapters/trafficflowtheory/theoryandconcepts/index.htm

Traffic Flow Parameters Capacity (cars per hour). The number of cars that pass a certain point during an hour. Car speed, (miles per hour, mph). In our simple model we will assume all cars are traveling at the same speed. Density (cars per mile). Suppose you took a snapshot of the highway from a helicopter and had previously marked two lines on the highway a mile apart. The number of cars you would count between these lines is the number of cars per mile. You can easily write down the interrelationship among these variables by using dimensionally consistent units: Capacity = Speed  Density [cars/hour] = [miles/hour]  [cars/mile]

Speed-Flow-Density Relationship Suppose you are flying in a helicopter, take the snapshot mentioned above, and count that there are 160 cars per mile. Suppose further that your first partner determines that the cars are crawling along at only 2.5 miles per hour. Suppose your second partner is standing by the road with a watch counting cars as described earlier. For one lane of traffic, how many cars per hour will your second partner count? Need: Capacity = ______ cars/hour. Know–How: You already know a relationship between mph and cars per mile Capacity = Speed × Density [cars/hour] = [miles/hr] × [cars/mile] Solve: Capacity = [2.5 miles/hr] × [160 cars/mile] = 400.cars/hour. 

The “follow rule” Number of car lengths between cars = [speed in mph]/[10 mph]

Use of rules Assume average length of a car is ~ 4.0 m on. We could pack about 400. of these vehicles/mile bumper-to-bumper The second equation assumes the follow rule of one spacing/10 mph

Effect of follow rule

Follow rule and max cars on road

Summary Kinematics is study of relationships among speed, distance, acceleration and time. For one-dimensional problems use a speed/time graph since the slope of the line gives acceleration and the area under the curve gives the distance travelled. For traffic analysis use: An empirical rule + capacity relationship: Capacity = Speed × Density [cars/hour] = [miles/hr] × [cars/mile]