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

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

Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience

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.