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Lyrx [107]
2 years ago
14

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?
Physics
1 answer:
elena-14-01-66 [18.8K]2 years ago
5 0

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.

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