Answer
$a = -2.16~m/s^2$
Work Step by Step
We can convert $20~km/h$ to units of $m/s$:
$(20~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 5.56~m/s$
We can find the time it takes the red car to reach $x = 44.5~m$:
$t = \frac{44.5~m}{5.56~m/s} = 8.0~s$
We can convert $40~km/h$ to units of $m/s$:
$(40~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 11.11~m/s$
We can find the time it takes the red car to reach $x = 76.6~m$:
$t = \frac{76.6~m}{11.11~m/s} = 6.9~s$
We can write an equation for the green car at $t = 8.0~s$:
$x-x_0 = v_0~t+\frac{1}{2}at^2$
$44.5-220 = 8.0~v_0+\frac{1}{2}a(8.0)^2$
$8.0~v_0+32~a+175.5 = 0$
We can write an equation for the green car at $t = 6.9~s$:
$x-x_0 = v_0~t+\frac{1}{2}at^2$
$76.6-220 = 6.9~v_0+\frac{1}{2}a(6.9)^2$
$6.9~v_0+23.8~a+143.4 = 0$
$(-\frac{32}{23.8})\times(6.9~v_0+23.8~a+143.4) = 0$
$-9.3~v_0-32~a-192.8 = 0$
We can add the two equations for the green car:
$-1.3~v_0-17.3 = 0$
$v_0 = -\frac{17.3}{1.3}$
$v_0 = -13.3~m/s$
$v_0 = (-13.3~m/s)(\frac{3600~s}{1~h})(\frac{1~km}{1000~m})$
$v_0 = -47.9~km/h$
We can find the acceleration of the green car:
$8.0~v_0+32~a+175.5 = 0$
$32~a = -175.5-8.0~v_0$
$a = \frac{-175.5-8.0~v_0}{32}$
$a = \frac{-175.5-(8.0)(-13.3)}{32}$
$a = -2.16~m/s^2$
