A particle is moving in a straight line is at a distance x from a fixed point in the straight line at time t seconds, where x = 2t^3 – 12t + 11. What is the velocity of the particle at the end of 2 seconds?
(a) 10 cm/sec
(b) 12 cm/sec
(c) 14 cm/sec
(d) 16 cm/sec
The question was posed to me during an interview.
The doubt is from Calculus Application topic in division Application of Calculus of Mathematics – Class 12
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Right choice is (b) 12 cm/sec
The best I can explain: We have, x = 2t^3 – 12t + 11 ……….(1)
Let v and f be the velocity and acceleration respectively of the particle at time t seconds.
Then, v = dx/dt = d(2t^3 – 12t + 11)/dt
= 6t^2 – 12 ……….(2)
Putting the value of t = 2 in (2),
Therefore, the displacement of the particle at the end of 2 seconds,
6t^2 – 12 = 6(2)^2 – 12
= 12 cm/sec.