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

Jul 26, 2018

It will take the object approximately $2.3 s$ to reach point B.

#### Explanation:

Begin the the equation

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

Which simplifies since ${v}_{i} = 0 \frac{m}{s}$

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

Finding the distance, use the formula for Euclidean distance:

$x = \sqrt{{\left(\textcolor{red}{3} - \textcolor{b l u e}{2}\right)}^{2} + {\left(\textcolor{red}{8} - \textcolor{b l u e}{9}\right)}^{2} + {\left(\textcolor{red}{8} - \textcolor{b l u e}{5}\right)}^{2}} m$

$x = \sqrt{11} m$

Substituting $x = \sqrt{11} m$ and $a = \frac{5}{4} \frac{m}{s} ^ 2$ in the formula:

$t = \sqrt{\frac{2 \sqrt{11} m}{\frac{5}{4} \frac{m}{s} ^ 2}}$

$t = \sqrt{\frac{8 \sqrt{11}}{5}} s$

$t \approx 2.3 s$