At the instant a traffic light turns green, a car starts from rest with a given constant acceleration of 0.5 m/s squared. just as the light turns green, a bus, traveling with a given constant speed 16.0 m/s, passes the car. the car speeds up and passes the bus sometime later. how far down the road has the car traveled when the car passes the bus?

d= 1024 m

Explanation:

Kinematics of the car

The car moves with uniformly accelerated movement we apply the following formula:

d= v₀t+ (1/2)*a*t² Formula (1)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

a: acceleration in m/s²

Data

v₀₁= 0   :initial speed of the car

a₁ =0.5 m/s² : acceleration of the car

Kinematics equation of the car

We replace data in the formula (1) :

d= 0+ (1/2)*0.5*t²

d= (0.25)*t²  Equation (1)

Kinematics of the bus

The car moves with uniformly movement ( constant speed) we apply the following formulas:

d= v*t   Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

Data

v =  16 m/s

Kinematics equation of the bus

We replace data in the formula (2) :

d= 16*t   Equation (2)

Problem development

when the car passes the bus the elapsed time (t) and the distance (d) is equal for both.

Equation (1) = Equation (2)=d

(0.25)*t² =16*t  We divide both sides of the equation by t

(0.25)*t = 16

t= 16/(0.25)

t= 64 s

We replace t in the equation  (2)

d= 16*t

d= 16*(64)

d= 1024 m

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)