(part 1 of 3) A blue car pulls away from a red stop-light just after it has turn

Question

(part 1 of 3)
A blue car pulls away from a red stop-light
just after it has turned green with a constant
acceleration of 0.5 m/s2. A green car arrives
at the position of the stop-light 5.5 s after the
light had turned green.
If t = 0 when the light turns green, at what
time does the green car catch the blue car if
the green car maintains the slowest constant
speed necessary to catch up to the blue car?
Answer in units of s.

(part 2 of 3)
What is the constant speed of the green car?
Answer in units of m/s.

(part 3 of 3)
What is the instantanous speed of the blue
car when the green car catches up to it?
Answer in units of m/s.

Explanation / Answer

let the green light occur at t=0
when blue car started with [u=0, a= 0.8 m/s^2]
--------------------------------------...
green car started at t=4, with v(slowest, constant)
=======================
let green car catch up with blue at [t=t]
so blue moved for (t) sec and green for (t-4)

s(blue) = 0 + 0.5 * 0.8 t^2 = 0.4 t^2
s(green) = v * (t -4)
for catching up > distances travelled by both = equal
v * (t -4) = 0.4 t^2
v = 0.4 t^2/[t-4]
v >> slowest
so dv/dt = 0, and d^v/dt^2 = +ve
dv/dt = 0.4[2t(t-4) - (1) t^2]/(t-4)^2 =0
[2t^2 - 8t - t^2] =0
t^2 = 8t >>>>>>> [ t= NOT zero, t = not 4]
t = 8 sec
this is the elapsed for blue, when green catches it
=================
v(slowest) [for t =8] = 0.4 *8^2/[8-4] = 6.4 m/s

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