# 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.

May 23, 2017

$3.31 \text{s}$

#### 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, ${v}_{0 x} = 0$

We'll make the initial position ${x}_{0}$ be $0$, and the position $x$ at time $t$ be $5.48 \text{m}$

The acceleration is constant at $1 \text{m"/"s}$

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

$x = {x}_{0} + {v}_{0 x} t + \frac{1}{2} {a}_{x} {t}^{2}$

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

$x = \frac{1}{2} {a}_{x} {t}^{2}$

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"