At the instant the traffic light turns green, a car starts with a constant acceleration of 3.00 ft/s^2. at the same instant a truck, traveling with a constant speed of 70.0 ft/s, overtakes and passes the car. how far from the starting point (in feet) will the car overtake the truck?

The car overtakes the truck at a distance d = 3266.2ft from the starting point

Explanation:

Problem Analysis

When car catches truck:

dc = dt = d

dc: car displacement

dt: truck displacement

tc = tt = t

tc: car time

tt : truck time

car kinematics :

car moves with uniformly accelerated movement:

d = vi*t + (1/2)a*t²

vi = 0 : initial speed

d = (1/2)*a*t² Equation (1)

Truck kinematics:

Truck moves with constant speed:

d = v*t Equation (2)

Data

We know that the acceleration of the car is 3.00 ft / s² and the speed of the truck is 70.0 ft / s .

Development problem

Since the distance traveled by the car is equal to the distance traveled by the truck and the time elapsed is the same for both, then we equate equations (1 ) and (2)

Equation (1) = Equation (2)

(1/2)*a*t² = v*t

(1/2)*3*t² = 70*t  (We divide both sides by t)

1.5*t = 70

t = 70 ÷ 1.5

t = 46.66 s

We replace t = 46.66 s in equation (2) to calculate d:

d = 70*46.66 = 3266.2ft

d = 3266.2 ft

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