Question

x axis is x = 4.00 - 7.00t2, 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

(a) 0 s

(b) 4.00 m

(c) -0.76 s

(d) +0.76 s

Explanation:

`x=4.00-7.00t^2`

It stops momentarily when the velocity,
`v` is 0.
`v` is the derivate of
`x`.

(a)
`v=(dx)/(dt)=-14.00t`

Setting this to 0,

`-14.00t=0`

`t=0`

(b) Substitute this value for
`t` in
`x` to get its position.

`x=4.00-7.00*0^2=4.00` m

It passes the origin when
`x=0`

`4.00-7.00t^2=0`

`7.00t^2=4`

`t^2=(4)/(7)`

`t=\pm\sqrt{(4)/(7)}`

(c) The negative time is
`t=-\sqrt{(4)/(7)} =-0.76 s`

(d) The positive time is
`t=+\sqrt{(4)/(7)} =+0.76 s`