One motor car A stands 24m in front of a motorcycle B. Both starts from rest along a straight road in the same direction. If A moves with uniform acceleration of 2 m/sec^2, then what will happen if B runs at a uniform velocity of 11 m/sec?

(a) A will not overtake B

(b) A will again overtake B and they will never meet again and again

(c) A will again overtake B and they will meet again

(d) A will again overtake B and they will never meet again

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My query is from Calculus Application topic in section Application of Calculus of Mathematics – Class 12

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Right choice is (d) A will again overtake B and they will never meet again

To explain: Let P and Q be the initial positions of the motor car and motorcycle B respectively, where PQ = 24m.

If possible, let us assume that B overtakes A after point R on the straight road after time t seconds from the start.

Then, considering the motion of motor car A, we get

PR = 1/2 (2) (t^2) = t^2

In this case when B runs at a uniform velocity 11 m/sec, we shall have, QR = 11t.

Therefore, in this case, QP + PR = QR gives,

Or 24 + t^2 = 11t

Or t^2 – 11t + 24 = 0

Or (t – 3)(t – 8) = 0

Or t = 3 or t = 8

Clearly, we are getting two real positive values of t.

Therefore, A and B will meet twice during the motion.

They will first meet after 3 seconds from start when the motorcycle B will overtake the motor car A.

But the velocity of A continuously increases, hence after a further period of (8 – 3) = 5 seconds A will again overtake B and they will never meet again.