A particle moves along a straight line such that its displacement at any timet is given by s = (t3 + 6t2 + 3t + 4)m. What is the velocity of the particle when its acceleration is zero?


Correct option is

– 9 m/s


As according to given problem,

               S = t3 – 6t+ 3+ 4

Instantaneous velocity

               v = ds/dt = 3t2 – 12+ 3                        …(1)

and acceleration


So acceleration will be zero when 6t – 12 = 0, i.e., t = 2 sec. And the so velocity when acceleration is zero, i.e., at  = 2 sec from Eqn. (1), will be

              v = 3 × 22 – 12 × 2 + 3 = – 9 m/s.



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A particle moving with velocity equal to 0.4 m/s is subjected to an acceleration of  0.15 m/s2 for 2 sec in a direction at right angle to its direction of motion. What is the magnitude of resultant velocity?


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