Distance, Speed, and Time Applications

Systems of Linear Equations Lesson #7: Distance, Speed, and Time Applications st Clans Rx. A student drove the 1245 km from Edmonton to Vancouver in 16/2 hours. This included a one hour stop in Golden and a 30 minute stop in Kamloops. She averaged 100 km/h on the divided highways and 75 km/h on the non-divided mountainous roads. How much time did she spend on the divided highways? driving time = 16호-1-½ = 15hours Distance (kim) speed (em/h) Time (h)) High way 100x 100 Mountainous Roads 75y 75 ५ x+y=1515-X 100275y=1245 CHECK 2-9 15 1000+7541245 LS (4.10.2) 25:100 (4.8)+75 (10.2) 15 PS: 15 LS = RS #1245 RS: 1245 LS-PS 100x+75 (15-x)=1245 y=15-2 100x+1125-75x-1245 y-15-4-8 25x = 120 X=4.8 y=10.2 she spent 4.8 hours on the divided highways Class Ex. 02 A small cruise boat took 3 hours to travel 36 km down a river with the current. On the return trip it took 4 hours against the current. Find the speed of the current and the speed of the small cruise boat in still water. Let xh be the speed of the boat in still water Let yam/h be the speed of the current (Distance(km) Speed (km/h) Time (n) Downstream 3(x+y) x+y 3 CHECK Upstream 4(x+y) 4 B(x+4) =36 4(x-4)=36 LS:3(10.5+1.5) LS: 4(10.5-1.5) =3(12)=36 RE:36 LSPS =4(9) 36. RS:35 LS=RS 3(x+y)=36 3x+3y=36(4) 4(x-4) = 36 subtracted b/c when going against the current (apstream) the speed will be lewer than going domstres 4x-4y=36 (13) 12x+12y=144 + 12x-12y = 108 24x = 252 x = 10.5 3x+3y=36 3005)-34=36 By 4.5 y=1.5 still water speed = 10.5 km/h Current speed = 1.5 FM/ Complete Assignment questions #1-#6 494 Systems of Linear Equations Lesson #7: Distance, Speed, and Time Applications Assignment In problems #1-#6 use the following procedure. a) Introduce variables to represent the unknown values. b) Form a system of equations involving the variables. c) Solve the system. d) Answer the problem and check the solution. 1. A cycle road test consists of a series of uphill and downhill sections. Li Na averaged 20 km/hr on the uphill sections and 40 km/hr on the downhill sections. If she completed the 35 km course in 1.3 hours, determine the length of the downhill sections. Ibistance (km) / Speed (m/h) Time (n) uphill 20x 20 404 40 x+y=1.3y = 1.3-X 20x +40 y = 35 Check: 2+4=1.3 20x+40=35 LS: 0.85 0.45 = 1.3 LS: 20(0.86)+40( PS:1.3 =35 RS: 35 LS=PS LS=RS 20x+40 (13-x)=35 y=1.3-x 20x+52-40x = 35 -20%--17 X0.85 Distance downhill = 40g * donet target This step! y=1.5-0.85 = 40(0.45) y = 0.45 = 18 km for distance! These are times! Question it gefing Length of downhill sections = 18km 2. A train travels 315 km in the same time that a car travels 265 km. If the train travels, on average, 20 km/h faster than the car, find the average speed of the car and the time taken to travel 265 km. Train Distance (Fm) speed (m/s) Time (n) 315 Car 265 ズ 215 y y x=y+20 312=265 315y = 265x CHECK 315 265 LS 315 2.5 (126) (106)= 2.5 LS RS LS: 126 RS: 265 x = y +20 RS: (106)+20=126 LS = PS 315y=265(y+20) x-y+20 315y=265y+5300 (106)+20 X = 126 50y = 5300 y = 106 (speed of car) Time taken 265-265 106 = 2.5 Time taken = 2.5 hours Average speed of car = 106Fm/h Systems of Linear Equations Lesson #7 Distance, Speed, and Time Applications 3. A small plane flying into a wind takes 3hr to travel the 780 km journey from Victoria to Prince Rupert. At the same time, a similar plane leaves Prince Rupert and reaches Victoria in 2hr. If the planes have the same cruising speed in windless conditions, determine the speed of the wind. Distance (km) speed ( Time (n) 3 Let x plane speed Let ym wind speed VPE 3(x-4) PRV 2.5(xry) x+y 2.5 3(x-y) = 980 3x-3y=780 (+5) 2.5(x+y)=780 2.52 +2.6y=780(16) 152-15y=3900 (add) 15x+15y=4680 30x8580 X = 286 CHECHI BX-34-780 2.5x+2.5y=7৪০ LS:3(286)-3(26) 15:2.5(286)+2.5(26) = 780 RS: 780 LS-FS = 480 RS: 760 LS=RS 3x-3y=780 3(286)-34 780 858-34 = 780 -39-78 y = 26 Wind speed is 26+m/n 4. A cyclist leaves home at 7:30 am to cycle to school 7 km away. He cycles at 10 km/h until he has a puncture; then he has to push his bicycle the rest of the way at 3 km/h. He arrives at school at 8:40 am. How far did he have to push his bicycle? Distance (rm) speed (%) 10 3 Time (h) 7:30AM 8:40AM Ihr. 10min.hr. Cycle /10 Push 9/3 x+y=7y=7-x (+=)(30)+3x+10=35 3x +104 =35 y=7-x 3x+10(9-x)-35 y=7-(s) 3x+70-10x35 -7x = -35 x=5 y=2 ! xy=7+= LS: (3)+(2) =7 RS: 7 LS RS 10 3 LS: + PS= LS-PS 6 6 He had to push his bicycle 2km