Answer:
(a) t=3.87 s :time at which Kathy overtakes Stan
(b) d=37.36 m
(c) vf₁ = 15.097 m/s : Stan's final speed
vf₂ = 19.31 m/s : Kathy's final speed
Explanation:
kinematic analysis
Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:
vf= v₀+at Formula (1)
vf²=v₀²+2*a*d Formula (2)
d= v₀t+ (1/2)*a*t² Formula (3)
Where:
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Nomenclature
d₁: Stan displacement
t₁ : Stan time
v₀₁: Stan initial speed
vf₁: Stan final speed
a₁: Stan acceleration
d₂: car displacement
t₂ : Kathy time
v₀₂: Kathy initial speed
vf₂: Kathy final speed
a₂: Kathy acceleration
Data
v₀₁ = 0
v₀₂ = 0
a₁ = 3.1 m/s²
a₂= 4.99 m/s²
t₁ = (t₂ +1) s
Problem development
By the time Kathy overtakes Stan, the two will have traveled the same distance:
d₁ = d₂
t₁ = (t₂ +1)
We aplpy the Formula (3)
d₁ = v₀₁t₁ + (1/2)*a₁*t₁²
d₁ = 0 + (1/2)*(3.1)*t₁²
d₁ = 1.55*t₁² ; Stan's cinematic equation 1
d₂ = v₀₂t₂ + (1/2)*a₂*t₂²
d₂ = 0 + (1/2)*(4.99)*t₂²
d₂ = 2.495* t₂² : Kathy's cinematic equation 2
d₁ = d₂
equation 1 = equation 2
1.55*t₁² = 2.495* t₂² , We replace t₁ = (t₂ +1)
1.55* (t₂ +1) ² = 2.495* t₂²
1.55* (t₂² +2t₂+1) = 2.495* t₂²
1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²
1.55t₂²+3.1t₂+1.55=2.495t₂²
(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0
0.905t₂² - 3.1t₂ - 1.55 = 0 Quadratic equation
Solving the quadratic equation we have:
(a) t₂ = 3.87 s : time at which Kathy overtakes Stan
(b) Distance in which Kathy catches Stan
we replace t₂ = 3.87 s in equation 2
d₂ = 2.495*( 3.87)²
d₂ = 37.36 m
(c) Speeds of both cars at the instant Kathy overtakes Stan
We apply the Formula (1)
vf₁= v₀₁+a₁t₁ t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s
vf₁= 0+3.1* 4.87
vf₁ = 15.097 m/s : Stan's final speed
vf₂ = v₀₂+a₂ t₂
vf₂ =0+4.99* 3.87
vf₂ = 19.31 m/s : Kathy's final speed
