The correct answer is (c) At t = 4
The explanation: Suppose, the two particle starts from rest at and move along the straight path OA.
Further assume that the distance between the particle is maximum after t minutes from start (before they meet again).
If we be the position of the particle after 30 minutes from the start which moves with uniform acceleration 5 m/min^2 and C that of the particle moving with uniform velocity 20 m/min, then we shall have,
OB = 1/2(5t^2) and OC = 20t
If the distance between the particle after t minutes from start be x m, then,
x = BC = OC – OB = 20t – (5/2)t^2
Now, dx/dt = 20 – 5t and d^2x/dt^2 = -5
For maximum or minimum values of x, we have,
dx/dt = 0
Or 20 – 5t = 0
Or t = 4
And [d^2y/dx^2] = -5 < 0
Thus, x is maximum at t = 4.