The correct option is C 24 m
Let the two cars be car A and car B.
For car A:
s1=ut+12at2⇒s1=4t+12×1×t2 (using second eqn of motion)
For car B:
s2=ut+12at2⇒s1=2t+12×2×t2 (using second eqn of motion)
On equating the distance, s1=s2
⇒4t+12×1×t2=2t+12×2×t2
⇒2t=t22
⇒t=4 s
Putting the value in any one of the equations, we get
s=4×4+12×1×42=24 m
Alternate sol:
Initial velocity of 1 w.r.t 2, u12=4−2=2 m/s
Acceleration of 1 w.r.t 2, a12=1−2=−1 m/s2
Finally, displacement of 1 w.r.t 2, S12=0
Using second equation of motion,
S12=u12+12a12t2
⇒0=2t−12×1×t2⇒t=4 s
∴ Distance travelled by 1st body =ut+12at2
=4×4+12×1×42=24 m