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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 secondI got this question during an online interview.The doubt is from Application of Derivative in section Application of Derivatives of Mathematics – Class 12

Answer»

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.


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