Exploring Engineering

Chapter 13

Engineering Kinematics

Kinematics and Traffic Flow

Transportation

Transportation 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

Distance vs. Time

0

10

20

30

40

50

60

70

80

0 1 2 3 4 5 6 7 8

Time in seconds

Dis

tan

ce i

n m

eter

s

Student Bus

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.

miles/hr.25hours 0.20

miles 0.5Speed Ave.

Kinematics 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

“Equations of motion”

Graphical View

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

Origin

Dis

tanc

e

Time

Stationary

Constant velocity

Variable velocity

Origin

Dis

tanc

e

Time

Stationary

Constant velocity

Variable velocity

speed

speed

Constant Velocity Algebraic Method

Apply Cartesian

geometry

Note: speed is distance/time Dimensions are

length/time Typical units ft/s, m/s

0

0

tt

xxV

)tV(txx 00

Origin

Dis

tanc

e

Time

x0

t0t

x

Constant VelocityCalculus Method

Speed is the rate of change of distance with time

We can integrate from the initial conditions (x0, t0) since V is constant

)t(txx constant tx 00 VV

Graphical Method

Consider the various regions in this graph Constant

speed Constant

acceleration Variable

acceleration

Origin

Spe

ed

Time

Constant speed, V0

Constant acceleration

Variable acceleration

Constant Acceleration Algebraic Method

0

0

tt

VVa

00 tta VV

00

00

21

21 Also,

ttaVV

VVVV

aver

aver

OriginS

peed

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

OriginS

peed

Time

Constant speed, V0

t0 t

V

Vaver

ExampleYou 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

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

0t sec

Spe

ed, f

t/s

a =

5.00

ft/s

147 ft/s

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 ContinuedS

peed

, mph

t, s

1.00 mile

15.

t1

Area = 1.00 miles = 15. t0or t0 = 1/15. miles = 4.0 minutes

Spe

ed, m

pht, s

2.00 mile

30.

t2

Area = 2.00 miles = 30. t0or t0 = 2.00/30. miles = 4.0 minutes

Sneaky Example MoralSo 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

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.

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

Subway ExampleS

peed

, ft/

s

6.0 ft/s2

-4.0 ft/s2

t1 t2t0= 0

73.

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

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.

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 ofone spacing/10 mph

Effect of follow rule

Speed, mph

Density, cars/mile

Capacity, cars/hr

0 400 0 10 200 2000 20 133 2667 30 100 3000 40 80 3200 50 67 3333 60 57 3429 70 50 3500 80 44 3556 90 40 3600 100 36 3636

Follow rule and max cars on road

0

500

1000

1500

2000

2500

3000

3500

4000

0 25 50 75 100

Speed, mph

Cap

acit

y, c

ars/

hr

0

50

100

150

200

250

300

350

400

0 20 40 60 80 100

Speed, mphD

ensi

ty, c

ars/

mil

e

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]