Question

A particle is moving in a straight line. The variation of position ‘x’ as a function of time ‘t’ is given as \(x = (t^3 – 6t^2 + 20t + 15) \,m\). The velocity of the body when its acceleration becomes zero is :

(1) 4 m/s

(2) 8 m/s

(3) 10 m/s

(4) 6 m/s

Answer 1

The correct option is (2) 8 m/s.

\(x = t^3 - 6t^2 + 20t + 15\)

\(\frac{dx}{dt} = v = 3t^2 - 12t + 20\)

\(\frac{dv}{dt} = a = 6t - 12\)

\(When\, a = 0\)

\(6t – 12 = 0; t = 2 sec\)

\(At \,t = 2 sec\)

\(v = 3(2)^2 - 12(2) + 20\)

\(v = 8\, m/s\)