# An object is at rest at (7 ,8 ,4 ) and constantly accelerates at a rate of 5/4 m/s^2 as it moves to point B. If point B is at (9 ,5 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jan 3, 2017

The answer is $= 2.45 s$

#### Explanation:

The distance betwwen A and B is

$= \sqrt{{\left(9 - 7\right)}^{2} + {\left(5 - 8\right)}^{2} + {\left(3 - 4\right)}^{2}}$

$= \sqrt{4 + 9 + 1}$

$= \sqrt{14} m$

We use the equation

$s = u t + \frac{1}{2} a {t}^{2}$

$u = 0$

So,

$\sqrt{14} = \frac{1}{2} \cdot \frac{5}{4} \cdot {t}^{2}$

${t}^{2} = \frac{8}{5} \sqrt{14} = 5.99 {s}^{2}$

$t = 2.45 s$