A particle is moving in one dimension (along x axis) under the action of a variable force. Its initial position was 16 m right of origin. The variation of its position (x) with time (t) is given as x = -3t3 + 18t2 + 16t, where x is in m and t is in s. The velocity of the particle when its acceleration becomes zero is m/s.
Answer
52
Solution
v = dxdt = -9t2 + 36t + 16
a = dvdt = -18t + 36 = 0 t = 2 s
v = -9(2)2 + 36(2) + 16 = 52 m/s
Answer: 52