Two particles start from rest from the same point and move along the same straight path; the first starts with a uniform velocity of 20 m/minute and the second with a uniform acceleration of 5 m/ min^2. Before they meet again, then what will be the maximum distance?

(a) 38m

(b) 36m

(c) 42m

(d) 40m

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Correct answer is (d) 40m

To elaborate: 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 ……….(1)

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.

Therefore, the maximum value is,

= 20*4 – (5/2)(4^2) [putting t = 4 in (1)]

= 40 m