Answer
(a) $x_1=(20.0~m/s)t+(1.25~m/s^2)t^2$
$x_2=1000~m-(30.0~m/s)t+(1.6~m/s^2)t^2$
(b) $t_1=23~s$
$t_2=120~s$
Work Step by Step
(a) $x_{10}=0$, $v_{10}=+20.0~m/s$, $a_1=+2.5~m/s^2$
$x_{20}=+1.0~km=+1000~m$, $v_{20}=-30.0~m/s$, $a_2=+3.2~m/s^2$
$x_1=x_{10}+v_{10}t+\frac{1}{2}a_1t^2$
$x_1=0+(20.0~m/s)t+\frac{1}{2}(2.5~m/s^2)t^2$
$x_1=(20.0~m/s)t+(1.25~m/s^2)t^2$
$x_2=x_{20}+v_{20}t+\frac{1}{2}a_2t^2$
$x_2=1000~m+(-30.0~m/s)t+\frac{1}{2}(3.2~m/s^2)t^2$
$x_2=1000~m-(30.0~m/s)t+(1.6~m/s^2)t^2$
(b) $x_1=x_2$
$(20.0~m/s)t+(1.25~m/s^2)t^2=1000~m-(30.0~m/s)t+(1.6~m/s^2)t^2$
$(0.35~m/s^2)t^2-(50.0~m/s)t+1000~m=0$
$t=\frac{-(-50.0~m/s)±\sqrt {(50.0~m/s)^2-4(0.35~m/s^2)(1000~m)}}{2(0.35~m/s^2)}$
$t_1=23~s$
$t_2=120~s$
