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?
I 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
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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