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, when will their distance be maximum?

(a) At t = 8

(b) At t = 6

(c) At t = 4

(d) At t = 2

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Question is taken from Calculus Application in chapter Application of Calculus of Mathematics – Class 12

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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.