A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position (x) with time (t) is given as x=-3 t³+18 t²+16 t, where x is in m and t is in s.
The velocity of the particle when its acceleration becomes zero is _________ m / s.

Solution

Physics sir · 3 steps

Differentiate twice: set the acceleration to zero to find the moment, then put it into the velocity.

1. Velocity. Differentiate the position once: v = dx/dt = -9t² + 36t + 16.
2. Acceleration. Differentiate again: a = dv/dt = -18t + 36. This is zero when t = 2 s.

   v(2) = -9(2)² + 36(2) + 16 = -36 + 72 + 16

3. So. The velocity when the acceleration becomes zero is 52 m s⁻¹. The starting position does not enter the calculation, because only derivatives of x are used.

The common slip: 80. Putting t = 2 s into x instead of v gives the position, 80, not the velocity.

The idea to keep. To find a quantity at the instant something else becomes zero, first solve for that instant from the condition, then put it into the quantity you want.

Answer from the NTA final answer key. Solution by PrepWiser.