Two cars start from rest at a red stop light. when the light turns green, both cars accelerate forward. the blue car accelerates uniformly at a rate of 4.3 m/s2 for 3.3 seconds. it then continues at a constant speed for 14.3 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 253.26 meters from where it started. the yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.1)how fast is the blue car going 2.6 seconds after it starts? how fast is the blue car going 7.2 seconds after it starts? how far does the blue car travel before its brakes are applied to slow down? what is the acceleration of the blue car once the brakes are applied? what is the total time the blue car is moving? what is the acceleration of the yellow car?

a)Blue car speed in t =2.6 s : v=11.8m/s

b)Blue car speed in t =7.2 s   : v=14.19m/s

c)Blue car distance : d=226.31m

d)acceleration of the blue car applying the brakes a= 3.6 m/s²

e)total time for the blue car: t=21.54s

f) acceleration of the yellow car: a=1.09 m/s²

Explanation:

Blue car kinematics:

Movement 1) time interval:(0,3.3)s: uniformly accelerated movement a=4.3m/s², v₀=0

d₀₋₁= v₀*t - ½ a*t²=0(3.3)*+½* 4.3*3.3²=23.41m

v₁=v₀+a*t=o+4.3*3.3

v₁=14.19m/s

Movement 2) time interval:(3.3,14.3)s Uniform movement

V is constant=v₁=14.19m/s

d₁₋₂=v*t=14.19*14.3=202.9m

Movement 3) applying the brakes : final speed=0

v₀=14.19m/s  ,v₃=0

d₂₋₃=253.26-22.41-202.9= 27.95m

v₃²=v₀₃²-2*a*d₂₋₃

0=14.19²-2*a*27.95

a=14.19²÷2*27.95

a=3.6m/s²

v₃=v₀-a*t₂₋₃

0=14.19-3.6t

t₂₋₃=14.19÷3.6

t₂₋₃=3.94s

yellow car kinematics:

d₀₋₃= v₀*t +½ a*t²

d₀₋₃=d₀₋₁+d₁₋₂+d₂₋₃=23.41m+202.9m+27.95m=254.26m

t=3.3+14.3+3.94=21.54 s

Problem development

a)Blue car speed in t =2.6 s  : t(0,3.3)s: Movement 1)

v₀₋₁=v₀+a*t=0+4.3*2.6

v(2.6s)=0+4.3*2.6

v(2.6s)=11.18m/s

b)Blue car speed in t =7.2 s  :t(3.3,14.3)s, Movement 2) : v is constant

v(7.2s)=14.19m/s

c)Blue car distance : d₀₋₂=Movement 1+Movement 2

d₀₋₂=d₀₋₁+d₁₋₂=23.41+202.9=226.31m

d)acceleration of the blue car applying the brakes :Movement 3)

a=3.6m/s²

e)total time for the blue car:

time(Movement 1)+time(Movement 2)+time(Movement 3)

t₀₋₃=3.3+14.3+3.94=21.54 s

f) acceleration of the yellow car

v₀=0

d₀₋₃=254.26m

t₀₋₃=21.54

d₀₋₃= v₀*t + ½ a*t²

254.26=0+ ½ *a*21.54²

a= 2*254.26 ÷ 21.54²

a=1.09 m/s²

c. newton's third law of motion