A car accelerates from rest at a constant rate of 2ms^ - 2 for some...

Using $v = u + atorv - u = at,$ we find that if $\left|\overset{→}{a}\right|$ is doubled,~$t$~will be halved.

If& $t$ is the time for accelerations, then $\dfrac{t}{2}$ is the time for retardation.

Now, $t + \dfrac{t}{2} = 3$

or~$\dfrac{3t}{2} = 3$

$t = 2s$

S = \dfrac{1}{2} \times 2 \times 2 \times 2 + \dfrac{1}{2} \times 4 \times 1 \times 1 = \left(4 + 2\right)m = 6m