Question

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.

Answer 1

`3.9` `"s"`

Explanation:

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.