At The Pass - Can Rose-Hulman's Solar Car Pass The University of Michigan's Solar Car in the alloted time? BRIEF ABSTRACT Given certain physical and legal restrictions can one car pass another in a given strip? GENERAL INFORMATION FileName: SOLARCAR Full title: At The Pass - Can Rose-Hulman's Solar Car Pass The University of Michigan's Solar Car in the allotted time? Developer: Kimberly Foltz, Mathematics and Computer Science Division, Indiana Academy for Science, Mathematics, & Humanities, Muncie IN 47306 USA Contact: Brian J. Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8412. Email: winkel@rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM In Sunrayce 95 , thirty-eight solar-powered cars competed in an endurance race from Indianapolis to Denver. Each solar car was accompanied by a lead vehicle and a chase vehicle to protect it as it traveled on interstates, highways, and other busy streets. For safety reasons when stopping, each driver in a solar car entourage allows a distance of 3 car lengths between the vehicles. Assume that the length of a car is 11 feet. Suppose on Day 7 of the race, the Solar Phantom built by Rose-Hulman Institute of Technology is in second place behind the entry from the University of Michigan (U of M) . The U of M solar car has had its power depleted to the point that it is traveling at the rate of 25 mph. The Rose entourage closes to within 3 car lengths of the U of Michigan chase vehicle and must settle in behind it at a speed of 25 mph. The Rose Solar Phantom has sufficient power at this point in the race to accelerate at a rate of 2 ft/second/second up to a maximum speed of 32 mph. When traffic permits, the Rose car decides to attempt a pass. In order to safely complete the pass, the entire entourage for the Solar Phantom must be able to pass the entire entourage for the U of M car and then pull back in leaving a safety zone of 3 car lengths between the Rose chase and the U of M lead. Suppose one of the rules of the race is that any solar car which intends to pass another car must complete the pass in 30 seconds so as not to obstruct traffic. Will the Solar Phantom be able to successfully complete the pass given the current conditions? KEYWORDS Position, velocity, acceleration,piecewise function. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Derivatives, antiderivatives. Time allotment - time management If given in class, it can be completed by small groups in one class period with perhaps some details left for homework. It can be given as a short-term assignment with a complete detailed solution due in two or three days. Expectations The problem is a fairly basic application of the relationships between position, velocity, acceleration, with just a slight complication. It might fit nicely as an extension to the standard text problems on these topics. Little guidance from the instructor should be necessary. Future payoffs Extensions References and Sources POSSIBLE SOLUTION(S) Note that each entourage is 9 car lengths long. Additionally, between the two entourages is a safety zone of 3 car lengths. Thus, relative to the Solar Phantom, the U of M car has an initial position of 132 feet. The U of M car has a constant velocity of 25 mph, so its position can be expressed as a linear function of the form s(t) = v0 t + s0. Converting the units to feet per second, mph2fps[x_]=x 5280 1/3600 22 x ---- 15 mph2fps[25] 110 --- 3 the position function for the U of M car is um[t_]:=110/3 t +132 The Solar Phantom position will be described in two different ways, depending upon whether it has reached its maximum velocity. The car starts with a constant acceleration of 2 f/s/s, which gives that velocity is vel[t_]=Integrate[2,t] + c c + 2 t Using the initial condition that the Solar Phantom is following the U of M car with an initial velocity of 25 mph, the constant of integration here is sol=Solve[vel[0]==110/3,c] c=c/.sol[[1]]; 110 {{c -> ---}} 3 This makes the velocity function v(t) = 2t + 110/3, which gives the position function as pos[t_]=Integrate[vel[t],t] + k 110 t 2 k + ----- + t 3 The initial position of the Solar Phantom relative to the U of M car is 0, so clearly k=0. k=0; pos[t] 110 t 2 ----- + t 3 rose1[t_]:=pos[t] We call this position function rose1(t) since it will be valid only until the velocity is 32 mph. This occurs at time t= sol=Solve[ vel[t]==mph2fps[32], t]; time=t/.sol[[1]]//N 5.13333 The Phantom can only accelerate until t=5.133 sec. After this time, the position of the car will be determined by a linear function with constant slope 32 mph. A point on this line is the point (5.133, pos(5.133) ). Substituting the coordinates of this point into the equation y = mx + b gives the y-intercept sol=Solve[pos[time]==mph2fps[32] * time + b,b]; yint=b/.sol[[1]] -26.3511 so that the position after t = 5.133 seconds is given by the function rose2[t_]=mph2fps[32] t + yint 704 t -26.3511 + ----- 15 The position of the Solar Phantom at any time t is described as a piecewise function composed of the two parts found above, split at the value of t where the velocity reached 32 mph. rose[t_]:=If[ t<=time, rose1[t], If[ t>time, rose2[t], ] ] From the graphs of each car's position over time, it is clear that eventually the Solar Phantom's position will overtake that of the U of M car. Plot[{um[t],rose[t]},{t,0,20}]; It is not as simple as finding where the graphs intersect to find when the pass is complete. This would put the two entourages side by side. The entire Rose entourage must move beyond the front of the U of M entourage, allowing an additional 3 car lengths before returning to the right lane of traffic. Thus, the Rose position must be 9+3 = 12 car lengths (132 feet) more than the U of M position. This will happen when t= 28.28 sec. sol=Solve[rose2[t]==um[t]+132,t]//N; tdone=t/.sol[[1]] 28.281 Note that this is verified by the fact that the positions at this time are exactly 132 feet apart. rose[tdone]-um[tdone] 132. Therefore, the pass will be completed in less than 30 seconds. No penalty and Rose takes the lead!!! ISSUES IN SOLUTION The velocity and position functions could have been obtained by using the Mathematica features for solving differential equations. For example: Clear[pos,vel,c] DSolve[vel'[t]==2,vel[t],t] {{vel[t] -> 2 t + C[1]}}