1. The problem statement, all variables and given/known data In a 100 m race between a lion and a tiger, the lion starts from rest and accelerates at a rate of 2 m/s2 for 5 seconds, then maintains a constant velocity for the rest of the race. The tiger starts from rest and accelerates at a rate 3 m/s2 for 4 seconds, then maintains a constant velocity for the rest of the race. Who wins the race? Support with evidence. At the moment that the winner crosses the finish line, how many meters behind is the loser? ∆d/d = displacement t= time Vi = initial velocity Vf = final velocity a= acceleration 2. Relevant equations Vavg = ∆d/∆t Vavg = (Vf + Vi)/2 a = (Vf- Vi)/t ∆d= (Vi)(t) + 1/2a(t^2) Vf^2 = Vi ^2 + 2ad 3. The attempt at a solution Part 1 Lion: Traveled 25 m in the first 5 seconds because: a = (Vf- Vi)/t 2 = (Vf- 0)/5 Vf = 10 Vavg = (Vf + Vi)/2 Vavg = 10 + 0 / 2 Vavg = 5 Vavg = ∆d/∆t 5 = ∆d/5 = 25 = ∆d Then it took 7.5 seconds to travel the remaining 75 m because it was going a constant speed of 10 m/s: Vavg = ∆d/∆t 10 = 75/t 10t =75 t= 7.5 7.5 seconds + 5 seconds = 12.5 Which means it took the lion 12.5 seconds to travel 100 m. Tiger: Traveled 24 m in the first 4 seconds because: a = (Vf- Vi)/t 3 = (Vf- 0)/4 Vf = 12 Vavg = (Vf + Vi)/2 Vavg = 12 + 0 / 2 Vavg = 6 Vavg = ∆d/∆t 6 = ∆d/4 = 24 = ∆d Then it took 6.3 seconds to travel the remaining 76 m because it was going a constant speed of 12 m/s: Vavg = ∆d/∆t 12 = 76/t 12t =76 t= 6.3 6.3 seconds + 4 seconds = 10.3 seconds Which means it took the tiger 10.3 seconds to travel 100 m. So the tiger won. Part 2: If it took 12.5 seconds for the Lion to go 100 m and 10.3 seconds for the tiger then the Lion still has 2.2 seconds to reach the finish line when the tiger goes across it. 12.5 seconds - 10.3 seconds = 2.2 seconds If the Lion is traveling at a rate of 10 m/s then: 10 m/s * 2.2 seconds = 22 m The lion is 22 meters behind the tiger when it crosses the finish line. Are these numbers correct?