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At the instant the traffic light turns green, a car starts with a const acceleration of 7.00 ft/sA2. At the same instant a truck, traveling wit constant speed of 70.0 ft/s, overtakes and passes the car. How far f the starting point (in feet) will the car overtake the truck?

User Shoji
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1 Answer

3 votes

Answer:

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

Step-by-step 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* = 70*t : we divide both sides of the equation by t

(1/2)*7*t = 70


t= (70*2)/(7)

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

User SolarLune
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