Answer:
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
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