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. When does the particle stop?

(a) 1/4 second

(b) 1/3 second

(c) 1 second

(d) 1/2 second

I got this question during an online interview.

The doubt is from Application of Derivative in section Application of Derivatives of Mathematics – Class 12

Answer 1

The correct choice is (c) 1 second

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

The particle stops when dx/dt = 0

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

3t^2 + 12t – 15 = 0

=>t^2 + 4t –5 = 0

=>(t – 1)(t + 5)= 0

Thus, t = 1 or t = -5

Hence, the particle stops at the end of 1 second.