Question

The position function x(t) of a particle moving along an x axis is x = 5.00 - 6.00t2, with x in meters and t in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin?

Answer 1

The position of the particle is given by, x(t) = 5.00-6.00: The velocity of the particle is given by, dt d(5.00-6.00r dt =-6.00(2t) =-12.0t (a) The particle stops moving when the velocity of the particle becomes zero and is given by, 0=-12.01 t= 0 =0 s the particle momentarily stops. Therefore, at time (b) The position of the particle when it stops is given by, x(0)=5.00 -6.00(0)* =5.00 m Therefore, the position of the particle at time t= 0 s is 5.00 m (c) At origin the position of the particle is zero and is given by, 0 = 5.00 – 6.00z 5.00 V6.00 = 0.913 s Therefore, at t=-0.913 s the particle passes through the origin. (d) At origin the position of the particle is zero and is given by, 0 = 5.00 – 6.00r 5.00 V6.00 = 0.913 s Therefore, at t= 0.913 s the particle passes through the origin.