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?

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

a women have faced negative attitudes in some areas , but other

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j