Question

A particle moving in a straight line covers a distance of x cm in t second, where x=ť+6t2-15t+18. What will be the velocity of the particle at the end of 2 seconds.



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Answer 1

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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Answer 2

please find the attachment