Question

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

Answer 1

`3.31"s"`

Explanation:

We're dealing with one-dimensional motion with constant acceleration.

What we can do first is find the displacement `s` during the time interval from point A to point B by using the three-dimensional distance formula:

`s = sqrt((3"m"-1"m")^2 + (1"m"-2"m")^2 + (4"m"-9"m")^2) = 5.48"m"`

Let's look at our known quantities:

Since the object is initially at rest, `v_(0x) = 0`

We'll make the initial position `x_0` be `0`, and the position `x` at time `t` be `5.48"m"`

The acceleration is constant at `1"m"/"s"`

To find the time duration `t` of the displacement `s`, we can use the equation

`x = x_0 + v_(0x)t + 1/2a_xt^2`

And since the initial velocity and position are both `0`, the equation becomes

`x = 1/2a_xt^2`

Here, the variable `x` is analogous to the displacement variable `s`, since they both represent the distance covered. Plugging in the known variables `a_x` and `x`, the time `t` elapsed during the displacement is

`5.48"m" = 1/2(1"m"/("s"^2))t^2`

`t = sqrt((2(5.48"m"/"s"))/(1"m"/("s"^2))) = color(red)(3.31"s"`