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Homework Statement
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?
t= time
Vi = initial velocity
Vf = final velocity
a= acceleration
Homework Equations
Vavg = ∆d/∆t
Vavg = (Vf + Vi)/2
a = (Vf- Vi)/t
∆d= (Vi)(t) + 1/2a(t^2)
Vf^2 = Vi ^2 + 2ad
The Attempt at a Solution
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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?