1. Exploring Engineering
Chapter 13
Engineering Kinematics

2. Kinematics and Traffic Flow

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

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

5. 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)

6. 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?

7. 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 m, or (Vb - Vs)(Δt) = 50. m
Solve: Solving for Δt we get
Δt = 50./(Vb - Vs) = 50./(10. – 2.5) = 6.7 seconds

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

9. 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.

10. 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

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

12. 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

13. 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

14. 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

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

16. 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

17. 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.

18. Example Continued a = 5.00 ft/s Speed, ft/s t sec
Need: x at V = 100. mph and t at 100 mph. Note: 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

19. 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!)

20. 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

21. 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!

22. 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 = 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 fps^2. For seated people this is increased to fps^2

23. 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 = ft/s2, V1(t1) = V2(0) = 50. mph = 73 ft/s
How: Speed/time graph
We know that x1 + x2 = 1320 ft

24. 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 fps^2. For seated people this is increased to fps^2
t0= 0 t1 t2

25. 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 = = 1,110 ft 
We know that x1 + x2 = 1320 ft 

26. 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

27. 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]

28. 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. 

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

30. 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

31. Effect of follow rule

32. Follow rule and max cars on road

33. 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]