Answers to Problem Set Number 02 for MIT (Spring 2008)

Transcription

1 Answers to Problem Set Number 02 for MIT (Spring 2008) Rodolfo R. Rosales (MIT, Math. Dept., room 2-337, Cambridge, MA 02139). March 10, Course TA: Timothy Nguyen, MIT, Dept. of Mathematics, Cambridge, MA Contents REGULAR PROBLEMS. 2 1 Haberman s book, problems & Haberman s book, problem Haberman s book, problem Haberman s book, problem Haberman s book, problem Haberman s book, problems & SPECIAL PROBLEMS Haberman s book, problem Haberman s book, problem Haberman s book, problem

2 MIT, Spring 2008 (Rosales). Answers to PS# 2. Haberman # & REGULAR PROBLEMS. 1 Haberman s book, problems & Statement for Many state laws say that, for each 10 m.p.h (16 k.p.h) of speed you should stay at least one car length behind the car in front of you. Assuming that people obey this law (i.e. exactly one car length), determine the density of cars as a function of speed (you may assume that the average length of a car is 16 feet (5 meters)). There is another law that gives a maximum speed limit (assume that this is 50 m.p.h (80 k.p.h)). Find the flow of cars as a function of density. Statement for The state laws on following distances, briefly discussed in exercise 61.1, were developed in order to prescribe spacing between cars such that rear-end collisions could be avoided. (a) Assume that the car immediately ahead stops instantaneously. How far would the driver following at u m.p.h travel, if (1) the driver s reaction time was τ, and (2) after then, the driver decelerated at constant maximum deceleration α? (b) The calculation in part (a) may seem somewhat conservative, since cars rarely stop instantaneously. Instead, assume that the first car also decelerates at the same maximum rate α, but the driver following still takes time τ to react. How far back does a car have to be, traveling at u m.p.h, in order to prevent a rear-end collision? (c) Show that the law described in exercise 61.1 corresponds to part (b), if human reaction time is about 1 second and the length of a car is about 16 feet (5 meters). Answers: These two problems are concerned with traffic laws that state that drivers must maintain a distance to the next car that is at least a minimum (stated in car lengths) which grows proportional to the car velocity. Assuming that the drivers follow this prescription exactly, it follows that d = Lu V, ( ) where d is the distance to the next car, L is the car length, u is the car velocity and V is the trigger velocity in the law, as in: State Law: Maintain a distance of one car length for each V increase in velocity. Typical numbers are V = 10 m.p.h = 16 k.p.h and L = 16 feet = 5 meters.

3 MIT, Spring 2008 (Rosales). Answers to PS# 2. Haberman # & Since then we end up having one car for every d + L distance, equation ( ) above leads to the velocity density relationship 1 (1 ρ = L + d = L + u ) ( ) 1 or u = V ρ L 1 V, ( ) and the corresponding flux law q = u ρ = ((1/L) ρ) V. Note that these laws give a constant wave speed c = dq = V and allow the car velocity to go to infinity as the density vanishes. This dρ is corrected by realizing that there is also a speed limit u M, in which case we get { ( ) } 1 u = min u M, ρ L 1 V. ( ) The critical density below which u = u M, is given by ρ c = applying for ρ ρ c or u u M. V, with equation ( ) L (V + u M ) Consider now the effects of the drivers reaction time τ, and the fact that cars do not change velocity instantaneously, but must have a finite acceleration. To simplify matters we will assume a constant deceleration = α. Thus a car will travel a distance D b = 1 u 2 2 α, ( ) from the moment the brakes are applied to the moment it stops. On the other hand, the distance traveled from the moment the driver sees that he must stop (say, the brake lights for the car ahead turn on) till the car actually stops, must include the driver s reaction time. Thus the total distance traveled is: D s = u τ + 1 u 2 2 α. ( ) It then follows that, in order to avoid a rear end collision, the distance between two cars traveling at speed u must be at least u τ. This then gives the following formula for the trigger velocity (see the note below) V = L τ. ( ) In particular, with L = 16 feet and τ = 1 second, we get a trigger velocity V = 16 feet per second, i.e. V = 10.9 m.p.h since a 1 mile = 5280 feet and an 1 hour = 3600 seconds. This trigger velocity is pretty close to the one used in many State Laws. Note: The calculation above is a bit sloppy, for it only checks that the cars are still one behind the other once they stop. What one should check is that one car remains behind the other always. However the car behind is always moving at a speed equal to or greater than the car ahead (for it starts slowing down at the same rate and starting from the same speed later). Thus the distance between the two cars is a non-increasing function of time. It follows that it is enough to check that it is positive once they stop, to know that it was always positive.

4 MIT, Spring 2008 (Rosales). Answers to PS# 2. Haberman # Haberman s book, problem If cars obey state laws on following distances (refer to exercise 61.1), what is the road capacity if the speed limit is 50 m.p.h (80 k.p.h.)? At what density and velocity does this maximum occur? Will increasing the speed limit increase the road s capacity? From the solution to problem 61.1 and { ( )} 1 u = min u M, V ρ L 1, ( ) { ( )} 1 q = u ρ = min ρ u M, V L ρ, ( ) where L = 16 ft = (1/330) mile is a car length, V = 10 m.p.h. is the trigger velocity in the law, and u M = 50 m.p.h. is the speed limit. Notice (see figure 2.1) that q is a tent-shaped, piece-wise Traffic flow laws q = q(ρ). Traffic flow laws q = q(u). Flux q. q = u M ρ (ρ m, q m ) q = - V ( ρ - 1/L) Flux q. q = V/L u = u M 0 Density ρ. ρ j = 1/L 0 Flow velocity u. Figure 2.1: Haberman s problem Left: flow-density curve for traffic flow that obeys (exactly) a typical state law. Right: flow-car velocity curve for the same situation. linear function of ρ, with slope u M for light traffic, and slope V for ρ for heavy traffic, with the maximum where these two straight lines intersect. Thus we have (we use the subscript m to indicate quantities associated with the road capacity) V ρ m = (u M + V ) L... For the numbers above ρ m = 55 c.p.m. q m = u M ρ m (road capacity)... For the numbers above q m = 2750 c.p.h. u m = u M... For the numbers above u m = 50 m.p.h.

5 MIT, Spring 2008 (Rosales). Answers to PS# 2. Haberman # Finally, we can write q = u M V L (u M + V ) = V L ( ) V 1, u M + V hence q m increases with u M. Another way to see this is to note that: In figure 2.1, the right hand side of the graph does not depend on u M, while the left side is a straight line through the origin, with slope u M. Since q m occurs at the intersection of these two lines, q m grows as u M does. Remark 2.1 One could do this problem by using u as the independent variable. That is, write 1 q as a function of u (not ρ), as follows: For 0 u u M, we have ρ = ( L 1 + u ), since the V distance between cars is L u V. Thus u q = u ρ = ( L 1 + u ), ( ) V which is clearly an increasing function of u (see the right frame in figure 2.1). Thus the maximum flow occurs at u = u M, etc. There is, however, one problem with this approach. Namely, u M corresponds not just to one value of the density, but to all the densities in the range 0 ρ ρ m = V L (u M + V ) all of which give the same u = u M value for the speed. For these values of the density, equation ( ) above does not apply! (On the left frame in figure 2.1, this is the left side of the tent ). However, it is easy to check directly (one must do this, if using this approach) that all the densities ρ < ρ m give a flux that is below the one given by q = q(u M ) in ( ). 3 Haberman s book, problem Referring to the theoretical flow-density relationship of exercise 71.1, show that the density wave velocity relative to a moving car is the same constant no matter what the density. In exercise 71.1 the car velocity u and wave velocity c are given by: Thus c u = a, u = a ln ( ) ρmax ρ and { ( ) } ρmax c = a ln 1. ρ and the wave velocity relative to the cars is a constant (all cars see the density waves approaching them, from the front, at the same speed.)

6 MIT, Spring 2008 (Rosales). Answers to PS# 2. Haberman # Haberman s book, problem Let c = dq dρ. Show that: c t + c c x = 0. ( ) Multiply ρ t + c ρ x = 0 by dc dρ, and use the chain rule: c t = dc dρ ρ t and c x = dc dρ ρ x. 5 Haberman s book, problem Show that if u = u(ρ) is determined by braking distance theory (see exercise 61.2), then the waiting time per car after a traffic light turns green is the same as the human reaction time for braking. Recall that (below the speed limit, which certainly applies in the circumstances of this problem) breaking distance theory gives the following density-flow velocity relationship ( ) 1 u ρ = V + 1 L, ( ) where V is the trigger velocity in the law (for each amount V in the car velocity, the separation between cars must augment by one car length) and L is the car length. The trigger velocity V is related to the reaction time τ by the equation V τ = L. This yields (for large enough densities that the cars velocity is below the speed limit) ( ) 1 q = L ρ V = (ρ j ρ) V, ( ) where ρ j = 1 L is the jamming density. Clearly, then: c(ρ j ) = V = L τ. ( ) From the solution to the red light turns green problem (see figure 5.2), a car at a distance d from the traffic light must wait a time t = d/c(ρ j ) before it starts moving. Thus τ W = waiting time per car = L c(ρ j ) where L is the car length and we used ( ) above. = τ = reaction time, ( )

7 MIT, Spring 2008 (Rosales). Answers to PS# 2. Haberman # & c(ρ) = x/t t x = c(ρ j ) t x = c(0) t ρ = ρ j ρ = 0 0 x Red Light Turns Green. Figure 5.2: Haberman s problem Characteristics in space-time for the red traffic light turns green problem. The critical characteristics are x = c(0) t and x = c(ρ j ) t, and the solution is (a) For x c(ρ j ) t, ρ = ρ j. (b) For c(ρ j ) t x c(0) t, ρ is given by the implicit equation c(ρ) = x/t. (c) For x > c(0) t, ρ = 0. 6 Haberman s book, problems & Assume that the car flow velocity is related to the car density by: u = u m ( 1 ρ ρ j ) = q = u m ( 1 ρ ρ j ) ( ρ = c = u m 1 2 ρ ) ρ j, ( ) where ρ j is the jamming density and u m is the car speed limit. Consider now the red light turns green problem and: First: Calculate the maximum acceleration of a car which starts approximately one car length behind a traffic light (i.e. x(0) = 1 ρ j ). Second: Calculate the velocity of a car at the moment it starts moving behind a light.

8 MIT, Spring 2008 (Rosales). Answers to PS# 2. Haberman # & The solution of the red light turns green problem is summarized in figure 6.3. Consider now the c(ρ) = x/t t x = c(ρ j ) t x = c(0) t ρ = ρ j ρ = 0 0 x Red Light Turns Green. Figure 6.3: Haberman s problems & Solution (in space time) of the red traffic light turns green problem, for the flow density relationship in ( ). In this case c(0) = u m, c(ρ j ) = u m, and ρ = 1 ( 2 ρ j 1 x ), for x u m t in the expansion fan region. t u m trajectory of a car starting at x = 1 ρ j (for t = 0). From exercise 63.6, the car s acceleration is given by α = ( ) 2 du ρ ρ x = dρ ( um ρ j ) 2 ρ ρ x. ( ) Remark 6.1 For completeness, we include the derivation of this last equation here. Car trajectories satisfy dx = u(ρ). Thus, using the chain rule: dt d 2 x dt = du dρ 2 dρ dt = du dρ (ρ t + u ρ x ) = du dρ (u c) ρ x. But c = d(ρ u) dρ = u + ρ du, from which ( ) follows. dρ

9 MIT, Spring 2008 (Rosales). Answers to PS# 2. Haberman # & It is clear then that the maximum acceleration will occur when ρ ρ x is the largest along a car trajectory. From the solution in figure 6.3, it should be clear that: as a car moves, ρ is nonincreasing along its trajectory. Further: the maximum slope (anywhere) of the solution decreases with time. Thus the maximum acceleration will occur right as the car starts moving and enters the expansion fan region. That is, the maximum acceleration will occur at a point x = u m t + 0, where ρ = ρ j and ρ x = ρ j. This yields: 2 t u m α max = 1 2 t m u m = 1 2 d u2 m, ( ) where t m = d u m is the time at which the car starts moving, and d is the initial distance from the car to the light. In particular, for d = 1 ρ j, this yields: α max = 1 2 ρ j u 2 m, ( ) Remark 6.2 Notice that equation ( ) gives an infinite acceleration for the first car behind the light when it goes green. This is, obviously, nonsense. It is generated by using formulas that assume a smooth distribution of cars, right at a place where there is a discontinuity. We should always be careful in situations like this, where we begin by making a continuum idealization for a nice car density function, and then we ask a question about a specific car right at the edge of a discontinuity, where the continuum approximation is doubtful. Of course, given what we say in the prior paragraph, we might ask if equation ( ) is reasonable (after all, one car length behind the light is still too small and below the continuum averaging distance). For typical numbers u m = 50 m.p.h and ρ j = 300 c.p.m, equation ( ) gives α max = m.p.h 2 = 10.4 m.p.h per second (possible for a racy car and driver, though a bit too high in general). At any rate, this gives an idea of how fast away from a discontinuity the continuum approximation becomes reasonable: a few car lengths is enough! Now consider the velocity of a car at the moment it starts moving after the light turns green. At that moment the local density is ρ j, so the velocity is u = 0, which is consistent with a finite acceleration and the fact that the car is starting from standing still. The only exception to this rule is the first car at x = 0, which right away starts moving with the speed that corresponds to ρ = 0, i.e.: u = u m. In the continuum theory small time intervals are neglected (in particular, the time it takes the first car to accelerate to maximum speed), and this leads to the rather absurd result of an infinite acceleration for the first car in the line behind the red light (see remark 6.2 above).

10 MIT, Spring 2008 (Rosales). Answers to PS# 1. Haberman # SPECIAL PROBLEMS. 7 Haberman s book, problem Assume that u(ρ) = u m (1 ρ/ρ j ), where u m is the speed limit and ρ j is the jamming density. For the initial conditions: ρ 0 for x < 0, ρ(x, 0) = ρ 0 (L x)/l for 0 x L, 0 for L < x, where 0 < ρ 0 < ρ j and 0 < L, determine and sketch ρ(x, t). ( ) ( d(ρ u) Note that c = c(ρ) = = u m 1 2 ρ ) a decreasing function of ρ. We now solve using dρ ρ j characteristics, as follows: Region (1) x < 0 at t = 0. Here ρ = ρ 0 along x = c 0 t + ζ, where ζ < 0 and c 0 = c(ρ 0 ), with c 0 < u m. Eliminating ζ, it follows that... ρ = ρ 0 for x < c 0 t. Region (2) 0 x L at t = 0. Here ρ = ρ ( ) 0 (L ζ) ρ0 (L ζ) along x = c t + ζ, L L where 0 ζ L. Eliminating ζ, it follows that... ρ = u m t + L x (u m c 0 ) t + L ρ 0 for c 0 t x u m t + L. Region (3) L < x at t = 0. Here ρ = 0 along x = u m t + ζ, where L < ζ. Eliminating ζ, it follows that... ρ = 0 for u m t + L < x. Summarizing, we have (see figure 7.4) ρ(x, t) = ρ 0 for x < c 0 t. u m t + L x (u m c 0 ) t + L ρ 0 for c 0 t x u m t + L. 0 for u m t + L < x. ( ) 8 Haberman s book, problem Assume that u(ρ) = u m (1 ρ 2 /ρ 2 j ), where u m is the speed limit and ρ j is the jamming density.

11 MIT, Spring 2008 (Rosales). Answers to PS# 1. Haberman # ρ(x, t) ρ 0 Solution, case c 0 < 0. 0 c 0 t u m t + L x Figure 7.4: Haberman s problem Solution to the initial value problem posed in equation ( ), for the case c(ρ 0 ) < 0, plotted for some arbitrary t > 0. The case c(ρ 0 ) > 0 is similar. For the initial conditions: ρ 0 for x < 0, ρ(x, 0) = ρ 0 (L x)/l for 0 < x < L, 0 for L < x, where 0 < ρ 0 < ρ j and 0 < L, determine and sketch ρ(x, t). d(ρ u) Note that c = c(ρ) = dρ characteristics, as follows: ( ) ( = u m 1 3 ) ρ2 a decreasing function of ρ. We now solve using ρ 2 j Region (1) x < 0 at t = 0. Here ρ = ρ 0 along x = c 0 t + ζ, where ζ < 0 and c 0 = c(ρ 0 ) with c 0 < u m. Eliminating ζ, it follows that... ρ = ρ 0 for x < c 0 t. Region (2) 0 x L at t = 0. Here ρ = ρ 0 (L ζ) L where 0 ζ L. Eliminating ζ, it follows that... ρ = L ρ 0 2 t (u m c 0 ) ( ) ρ0 (L ζ) along x = c t + ζ, L t (u m c 0 ) λ, L 2

12 MIT, Spring 2008 (Rosales). Answers to PS# 1. Haberman # where λ = u m t + L x, and c 0 t x u m t + L. Region (3) L < x at t = 0. Here ρ = 0 along x = u m t + ζ, where L < ζ. Eliminating ζ, it follows that... ρ = 0 for u m t + L < x. Summarizing, we have (see figure 8.5) ρ 0 for x < c 0 t, L ρ 0 ρ(x, t) = t (u m c 0 ) λ for c 2 t (u m c 0 ) L 2 0 t x u m t + L, 0 for u m t + L < x, ( ) ρ(x, t) ρ 0 Solution, case c 0 < 0. 0 c 0 t u m t + L x Figure 8.5: Haberman s problem Solution the the initial value problem posed in equation ( ), for c(ρ 0 ) < 0, plotted for some arbitrary t > 0. The case c(ρ 0 ) > 0 is similar. 9 Haberman s book, problem Consider the (non-dimensionalized) partial differential equation ρ t ρ 2 ρ x = 0, < x < and t > 0. ( )

13 MIT, Spring 2008 (Rosales). Answers to PS# 1. Haberman # (a) Why can t this equation model a traffic flow problem? (b) Solve this P.D.E. by the method of characteristics, subject to the initial conditions: 1 for x < 0, ρ(x, 0) = 1 x for 0 x 1, ( ) 0 for 1 < x. (a) If equation ( ) was a model for traffic flow, with ρ the car density, then the flow would have to be: q = C ρ 3 /3, where C is some constant. However, q(0) = 0, so that C = 0. But then q = ρ 3 /3, which is always negative (thus not acceptable as a model for traffic flow). (b) The characteristic equations are dx dt = ρ2 and dρ = 0. ( ) dt For the given initial conditions this yields (see figure 9.6): t (3) (2) (1) ζ x = - t + ζ ρ = 1 x = - t 0 ζ 1 x x = - (1 - ζ 2 ) t + ζ ρ = 1 - ζ ρ = 0 Characteristics. Figure 9.6: Haberman s problem Characteristics for the initial value problem given by equations ( ). Note that the characteristics neither cross, nor do they leave any gaps so that we do not have to worry about introducing shocks or expansion fans into the solution.

14 MIT, Spring 2008 (Rosales). Answers to PS# 1. Haberman # Region (1) 1 < x at t = 0. Here ρ = 0 along x = ζ, for ζ > 1. Thus ρ = 0 for x > 1. Region (2) 0 x 1 at t = 0. Here ρ = 1 ζ along x = (1 ζ) 2 t + ζ, for 0 ζ 1. Eliminating ζ from these last two equations, we get ρ 2 t + ρ + x 1 = 0, for t x 1. Therefore... ρ = t + 4 t + 1 x for t x 1. 2 t Note that the positive square root gives the right behavior as t 0, with no singularity. The other root is spurious (introduced when eliminating ζ above). Region (3) x < 0 at t = 0. Here ρ = 1 along x = t + ζ, for ζ < 0. Therefore... To summarize, the solution is: 1 for x < t, ρ(x, 0) = t + 4 t + 1 x for t x 1, 2 t 0 for 1 < x. ρ = 1 for x < t. ( ) THE END.