Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.1 m/s2 while Kathy maintains an acceleration of 4.99 m/s. 2 (a) Find the time at which Kathy overtakes Stan. s from the time Kathy started driving (b) Find the distance she travels before she catches him (c) Find the speeds of both cars at the instant she overtakes him. Kathy m/s Stan m/s

(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