Given:
Position of particle, x = At² – Bt².
Numerical value of coefficient , A, A = 3.0 m/s²
Numerical value of coefficient , B , B = 1.0/s³
For the equation given as x = At² – Bt³ to be dimensionally correct, the dimensions on the left side must be equal to the right. However this is true if the dimensions of At² and Bt3 matches that of x.
Since x is measured in meters in S.I units, therefore the dimensions of At² and Bt³ should also be meters.
The dimensions of t in S.I units is seconds, therefore the dimension of A must be meters/ second². Thus one can see that the multiplication of A and t² will bring a quantity with dimensions of .
Therefore the dimensions of A is meters/second².
Similarly, if the dimension of B is meters/second³, then on multiplication with t³ it will produce a quantity with dimensions of .
Therefore the dimension of B is meters/second³

Therefore we find that the last time the object was at positive x axis was just before 3s.
One can substitute various value of time t to see the behavior of the object. It is clear from the equation that the particle starts at_x = 0 m, moves towards the positive x axis with time. The maximum value of x occurs at t = 2 sand is given as:

The particle has moved 4 m in the first 2 seconds, then reversed its direction and moved along the negative x axis. At t = 3 s, the particle managed to reached the position from where it started its motion.Therefore the total distance travelled by the particle till 3 seconds is twice the maximum displacement towards positive x axis that is 2(4 m) = 8 m.
The position of the particle at t = 4 s is given as:

The particle has went to x = 16 m towards the negative x axis from t = 3 s to t = 4s. Therefore the total distance travelled by the particle till 4 seconds is the sum of the distance travelled in first 3 seconds, which is and the distance travelled by it in the next second, which is 16m.

Thus the path length of the particle in first 4 seconds is 26 m.
(d) The path length in the part above accounts for the total distance travelled by the particle in first 4 seconds. However to account for its displacement, one has to calculate the particle position relative to its initial point.
As said above, the particle reaches its initial position at t = 3 s, therefore the net displacement of the particle is calculated from the particle position at t = 4 s which is -16 m.
Therefore the net displacement of the particle in first four seconds is - 16m.
(e) The velocity (say v) of the particle at any instant is given by differentiation of its position x with respect to time as:






a = 2A – 6Bt ….….….….….….2