Question

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

Answer 1

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.