As a car travels down a highway, it travels a constant speed of 70 miles per hour. It burns fuel as it moves. What happens to the mass of the car as the fuel is burned? How does the affect the kinetic energy of the car?

d= 1400 ft :Distance from the starting point where the car reaches the truck

Explanation:

Car Kinematics : the car has uniformly accelerated movement, so we apply the following equation for distance

d₁= v₀₁*t + (1/2)* a₁*t₁² Equation 1

d₁: car distance (ft)

v₀₁: car initial speed (ft/s)

a₁ : car acceleration  (ft/s²)

t1 :  car time (s)

Truck Kinematics: The truck moves with uniform motion (constant speed), so the equation for distance is:

d₂= v₂*t₂ Equation (2)

d₂: truck distance (ft)

v₂: truck  speed (ft/s)

t₂ :  truck time (s)

Known data

v₀₁ = 0

a₁ = 7 ft/s²

v₂ = 70 ft/s

Problem development

The time and distance for when the car catches the truck is the same, then:

d₁=d₂   and  t₁=t₂= t

Equation (1  )= the Equation ( 2)

v₀₁*t₁ + (1/2)* a₁*t₁²  =  v₂*t₂

v₀₁*t + (1/2)* a₁*t²=  v₂*t

0*t + (1/2)*7*t² = 70*t   : we divide both sides of the equation by t

(1/2)*7*t = 70

t= 20 s

We replace t= 20 s in the equation (2)

d= 70*20 = 1400 ft :Distance from the starting point where the car reaches the truck