- #1

- 10

- 0

## 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

[/B]

*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?

Are these numbers correct?