12. A particle moves along a straight line such that its displacement at any time t is given by s = ( t3 − 6t2 + 3t + 4 ) metres The velocity when the acceleration is zero is

12. A particle moves along a straight line such that its displacement at any time t is given by s = \( ( t^3 − 6t^2 + 3t + 4 ) \) metres The velocity when the acceleration is zero is

(a)       \(3 \text{ } ms^{−1} \)

(b)  \(−12 \text{ } ms^{−1} \)

(c)      \(42 \text{ } ms^{−2} \)

(d)   \(−9 \text{ } ms^{−1} \)

Answer :  (d)   \(−9 \text{ } ms^{−1} \)

\( s = t^3 - 6t^2 + 3t + 4 \)

\( v =\)\(\frac{ds}{dt}\)\( = 3t^2 - 6(2)t + 3(1)t^0 + 0 \)

\(  v = 3t^2 - 12t + 3 \)

\( a = \)\(\frac{dv}{dt}\)\( = \)\(\frac{d^2 s}{dt^2} \)\( = 3(2)t - 12 \)

\( \text{When t = 2 ,      a = 0} \)

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

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

\( = 12 - 24 + 3 \)

\(−9 \text{ } ms^{−1} \)