Question

The position function x(t) of a particle moving along an x axis is x = 4.00 - 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 function x(t) of a particle moving along an x axis is $x=4.00 - 6.00t^2$

a) The point at which particle stop, it's velocity = 0 m/s

  So dx/dt = 0

        0 = 0- 12t = -12t

  So when time t= 0, velocity = 0 m/s

    So the particle is starting from rest.

At t = 0 the particle is (momentarily) stop

b) When t = 0

 $x=4.00 - 6.00*0^2 = 4m$

SO at x = 4m the particle is (momentarily) stop

c) We have $x=4.00 - 6.00t^2$

   At origin x = 0

  Substituting

         $0 = 4.00 - 6.00t^2\\ \\ t^2 = \frac{2}{3}$

         t = 0.816 seconds or t = - 0.816 seconds

So when  t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.