Answer
(a) Car A will be ahead of car B just after cars starts to move.
(b) Both the cars are at same point at time $ t=0,\space 2.20\space s$ and $5.73\space s$
(c) Distance between A and B is not changing at $t=1.00\space s$ and $4.34\space s$
(d) Acceleration of A and B will be same at $t=2.67\space s$
Work Step by Step
(a) Velocity of car A at time t, $v_A=\frac{dx_A(t)}{dt}=\alpha+2\beta t$
Velocity of car B at time t, $v_B=\frac{dx_B(t)}{dt}=2\gamma t-3\delta t^2$
At $t=0,\space v_A(t=0)=\alpha=2.60\space m/s$ and $ v_B(t=0)=0$
Since initially speed of car A is greater than that of car B, so car A will be ahead of car B just after cars starts to move.
(b) For both the cars to be at same point,
$x_A=x_B\Rightarrow \alpha t+\beta t^2=\gamma t^2-\delta t^3$
One solution of this equation is $t=0$, which is starting point.
To find other solutions, $\delta t^2+(\beta-\gamma)t+ t++\alpha=\gamma t^2$
$\Rightarrow t=\frac{-(\beta-\gamma)\pm\sqrt{(\beta-\gamma)^2-4\alpha\delta}}{2\delta}=\frac{-(1.20-2.80)\pm\sqrt{(1.20-2.80)^2-4\times2.60\times0.20}}{2\times0.20}=(4.00\pm1.73)\space s$
$\Rightarrow t=2.20\space s$ and $t=5.73\space s$
So both the cars are at same point at time $ t=0,\space 2.20\space s$ and $5.73\space s$
(c) The instant at which distance between A and B is not changing, $\frac{d(x_A-x_B)}{dt}=0$
$\Rightarrow \frac{dx_A}{dt}-\frac{dx_A}{dt}=(\alpha+2\beta t)-(2\gamma t-3\delta t^2)=0$
$\Rightarrow 3\delta t^2+2(\beta-\gamma)+\alpha=0$
$\Rightarrow t=\frac{-2(\beta-\gamma)\pm\sqrt{4(\beta-\gamma)^2-12\alpha\delta}}{6\delta}=\frac{-2(1.20-2.80)\pm\sqrt{4(1.20-2.80)^2-12\times2.60\times0.20}}{6\times0.20}$
$\Rightarrow t=(2.67\pm1.67)\space s=1.00\space s$ and $4.34\space s$
So, distance between A and B is not changing at $t=1.00\space s$ and $4.34\space s$
(d) Acceleration of car A at time t, $a_A=\frac{dv_A(t)}{dt}=2\beta$
Acceleration of car B at time t, $a_B=\frac{dv_B(t)}{dt}=2\gamma-6\delta t$
So, for acceleration of A and B to be same,
$a_A=a_B\Rightarrow 2\beta=2\gamma-6\delta t$
$\Rightarrow t= \frac {2(\gamma-\beta)}{6\delta}=\frac {2(2.80-1.20)}{6\times0.20}=2.67\space s$
So, acceleration of A and B will b same at $t=2.67\space s$
