# 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.

Jun 3, 2017

$3.9$ $\text{s}$

#### Explanation:

We're asked to find the time it takes for the object to move from $\left(3 , 5 , 1\right)$ to $\left(9 , 9 , 8\right)$ 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

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

Our known quantities are

• $\Delta x$ 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:

$\Delta x = \sqrt{{\left(9 \text{m"-3"m")^2 + (9"m"-5"m")^2 + (8"m"-1"m}\right)}^{2}}$

= color(red)(10.0"m"

• ${v}_{0 x}$ 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 $\left(3 , 5 , 1\right)$ to $\left(9 , 9 , 8\right)$ in $3.9$ seconds.