Answer:
Time it takes for player 1 to catch player 2: t = 7,55s
The players traveled a distance of: d = 34.2m
Explanation:
Problem Analysis
When player 1 catches player 2 they will both have traveled the same distance d and player 2 will have spent 1 second more time, then:
d₁ = d₂ = d
t₂ = t₁ + 1
Player kinematics 1
Player 1 moves with uniformly accelerated movement:
[tex]d = v_i*t+\frac{1}{2} a*t_1^2[/tex] (Formula 1)
d: displacement in meters (m)
vi: initial speed = 0
a: acceleration = 0.12m/s²
t₂ = time in s
We replace data in formula (1)
d = (1/2)(0,12)×t₁²
d = 0,6t₁² equation (1)
Player kinematics 2
Player 2 moves with constant speed:
d = v × t₂ Formula (2)
d: displacement in meters (m)
v = speed in m/s = 4 m/s
t₂: time in s
t₂ = t₁+1
We replace data in formula (2)
d = 4 × (t₁+1)
d = 4t₁ + 4 equation (2)
equation (1) = equation (2)
0,6t₁² = 4t₁ + 4
0,6t₁² - 4t₁ - 4 = 0
Applying the quadratic formula:
a = 0,6
b = -4
c = -4
[tex]t_1 = \frac{-b+\sqrt{b^2-4(a)(c)} }{2a}[/tex]
[tex]t_1' = \frac{-(-4)+\sqrt{(-4)^2-4(0,6)(-4)} }{2(0,6)}=7,55s[/tex]
[tex]t_1'' = \frac{-(-4)-\sqrt{(-4)^2-4(0,6)(-4)} }{2(0,6)}=-0.88s[/tex]
We take value of the positive time because the negative time does not exist
t₁=7,55s
We replace in the equation (1) to calculate the distance traveled
d = 0,6(7,55)² = 34.2m
