Question

A particle moving in a straight line covers a distance of x cm in t second, where x = t^3 + 6t^2 – 15t + 18. What will be the velocity of the particle at the end of 2 seconds?

(a) 20cm/sec

(b) 22cm/sec

(c) 21cm/sec

(d) 23cm/sec

This question was posed to me in class test.

Query is from Application of Derivative in division Application of Derivatives of Mathematics – Class 12

Answer 1

Correct option is (c) 21cm/sec

The explanation is: We have, x = t^3 + 6t^2 – 15t + 18

Let, v be the velocity of the particle at the end of t seconds. Then,

v = dx/dt = d/dt(t^3 + 6t^2 – 15t + 18)

So, v = 3t^2 + 12t – 15

Thus, velocity of the particle at the end of 2 seconds is,

[dx/dt]t = 2 = 3(2)^2 + 12(2) – 15 = 21cm/sec.