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A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the described by the particle during the 10^th second of its motion?(a) 38.5cm(b) 37.5cm(c) 38cm(d) 39.5cmI had been asked this question during an interview.This is a very interesting question from Calculus Application in section Application of Calculus of Mathematics – Class 12

Answer» CORRECT option is (a) 38.5cm

The best explanation: LET, the particle moving with a uniform acceleration of F cm/sec^2.

By question initial velocity of the particle is U = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.

Thus, using the formula v = u + ft we get

34 = 10 + f*8

Or 8f = 24

Or f = 3

Therefore, the required acceleration of the particle is 3 cm/sec^2.

The space described during the 10TH second of its motion is,

= [10 + 1/2(3)(2*10 – 1)][using the formula st = ut +1/2(f)(2t – 1)]

= 10 + 28.5

= 38.5cm.


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