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 and B runs at a uniform velocity of 11 m/sec then when will they meet?

(a) After 3 seconds from start when the motorcycle B will overtake the motor car A

(b) After 4 seconds from start when the motorcycle B will overtake the motor car A

(c) After 2 seconds from start when the motorcycle B will overtake the motor car A

(d) After 6 seconds from start when the motorcycle B will overtake the motor car A

The question was asked in exam.

The above asked question is from Calculus Application in portion Application of Calculus of Mathematics – Class 12

NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options

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Right option is (a) After 3 seconds from start when the motorcycle B will overtake the motor car A

The explanation: 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 ata uniform velocity of 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.