An object is at rest at `(3 ,5 ,1 )` and constantly accelerates at a rate of `4/3 m/s^2` as it moves to point B. If point B is at `(9 ,9 ,8 )`, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

We're asked to find the time it takes for the object to move from `(3, 5, 1)` to `(9, 9, 8)` at a constant acceleration of `4/3"m"/("s"^2)`. (This is one-dimensional motion, as it's moving in a straight line, regardless of the fact that the coordinates are three-dimensional.)

To do this, we can use the equation

`Deltax = v_(0x)t + 1/2a_xt^2`

Our known quantities are

• `Deltax` is the change in position of the object from the two coordinate points. We can find the distance between these two coordinates and call this the change in position:

`Deltax = sqrt((9"m"-3"m")^2 + (9"m"-5"m")^2 + (8"m"-1"m")^2)`

`= color(red)(10.0"m"`

• `v_(0x)` is the initial velocity of the object. Since it's originally at rest, this value is zero.

• `a_x` is the object's (constant) acceleration, which is `4/3"m"/("s"^2)`

Plugging in our known values, we have

`color(red)(10.0"m") = (0)t + 1/2(4/3"m"/("s"^2))t^2`

`color(red)(10.0"m") = (2/3"m"/("s"^2))t^2`

`t = sqrt((10.0"m")/(2/3"m"/("s"^2))) = color(blue)(3.9` `color(blue)("s"`

The object will travel the distance from `(3, 5, 1)` to `(9, 9, 8)` in `3.9` seconds.