Two cars start off to race with velocities $4\text{m/s}$ and $2\text{m/s}$ and travel in a straight line with uniform accelerations $1{\text{m/s}}^2$ and $2{\text{m/s}}^2$ respectively. If they reach the final point at the same time instant, then what is the length of the path?

The correct option is C $24\text{m}$
Let the two cars be car $A$ and car $B$.
For car $A$:
$s_1=ut+\frac{1}{2}a{t}^2\Rightarrow s_1=4t+\frac{1}{2}\times 1\times t^2$ (using second equation of motion)
For car $B$:
$s_2=ut+\frac{1}{2}a{t}^2\Rightarrow s_2=2t+\frac{1}{2}\times 2\times t^2$ (using second equation of motion)
On equating the distance, $s_1=s_2$
$\Rightarrow 4t+\frac{1}{2}\times 1\times t^2=2t+\frac{1}{2}\times 2\times t^2$
$\Rightarrow 2t=\frac{t^2}{2}$
$\Rightarrow t=4\text{s}$
Putting the value in any one of the equations, we get
$s=4\times 4+\frac{1}{2}\times 1\times 4^2=24\text{m}$
Alternate Solution:
Initial velocity of $1^{\text{st}}$ car w.r.t. 2, $u_{12}=4-2=2\text{m/s}$
Acceleration of $1^{\text{st}}$ car w.r.t. 2, $a_{12}=1-2=-1{\text{m/s}}^2$
Finally, displacement of 1^st car w.r.t 2, $S_{12}=0$
Using second equation of motion,
$S_{12}=u_{12}+\frac{1}{2}{a}_{12}{t}^2$
$\Rightarrow 0=2t-\frac{1}{2}\times 1\times t^2\Rightarrow t=4\text{s}$
$\therefore$ Distance travelled by 1^st car $=ut+\frac{1}{2}a{t}^2$
$=4\times 4+\frac{1}{2}\times 1\times 4^2=24\text{m}$