There are two checkpoints on a road namely Pt. A and Pt. B. There are 5 persons standing at the Pt. A namely P , Q , R , S, T. They have a car such that P , Q , R , S , T can travel to Pt. B in 1 , 3 , 6 , 8 , 12 seconds respectively. If only two persons can get into the car and person with more driving time drives also one person has to bring car back to Pt. A unless all have arrived Pt. B. How can you transport all the persons to Pt. B in 30 seconds or less.
P and Q go, Q returns back alone. 6 seconds passed
S and T go. P returns back alone. 13 seconds passed.
P and R go, P returns alone. 7 seconds passed.
P and Q go. 3 seconds passed.
Total time: 29 seconds
No real-world car can accelerate faster than $5g$, so the maximum acceleration of the car is under $50m/s^2$. Even if the car instantly decelerates (against a wall or something), the fact that $P$ can do the trip in $1$ second means that $A$ and $B$ are at most $25m$ apart (more realistically $1m$), so running this distance takes less than five seconds at a very generous estimate.
Solution: $P$ and $Q$ use the car, and $R, S, T$ run. Total time: under $5$ seconds.
EDIT to merge the comments: Suppose that some of the people have much more trouble running, since they also seem to have trouble driving a car. In that case, we should also consider an average $0.4g$ car and not a $5g$ dragster. In this case, the maximum distance achievable in $1$ second is $2m$, so let's assume our car is placed sideways in a dark alley to ensure proper lithobraking.
Now let's look at our people. They may have more or less trouble walking, or getting into cars, or driving, but what we do know is that there is a lower bound on how fast they can get into and out of cars. Suppose they have to move $1m$ to enter the car, and $1m$ to leave again. Entering and leaving the car is part of driving the car from $A$ to $B$, so any person can therefore move at least $2m$ by foot in the time he needs to move $2m$ by car.
The optimal solution is now as follows: Everybody walks, and the car stays there. This takes at most $12$ seconds, potentially much less if the given people have much more trouble operating cars than walking.
1 and 3 go. (3 seconds)
1 comes back. (4 seconds)
12 and 8 go (16 seconds)
3 comes back. (19 seconds)
1 and 6 go (25 seconds)
1 comes back (26 seconds)
1 and 3 go (29 seconds)
T and P gets in; T drives to pt.B and P drives back to pt. A(they now have 17secs left)
S gets in with P still inside; S drives to pt. B then P drives back (8secs left)
R gets in then drives to pt.B, P drives back (1sec remaining)
Q gets in then P drives them both to pt.B (0sec remaining)
Transporting them all took exactly 30secs