# An object is at rest at (8 ,2 ,5 ) and constantly accelerates at a rate of 7/4 m/s^2 as it moves to point B. If point B is at (6 ,5 ,1 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

time $t = 2.480821248 \text{ }$seconds

#### Explanation:

Given object at rest
${v}_{0} = 0 \text{ }$initial velocity

distance traveled by the object
$s = \sqrt{{\left({x}_{B} - {x}_{A}\right)}^{2} + {\left({y}_{B} - {y}_{A}\right)}^{2} + {\left({z}_{B} - {z}_{A}\right)}^{2}}$
$s = \sqrt{{\left(6 - 8\right)}^{2} + {\left(5 - 2\right)}^{2} + {\left(1 - 5\right)}^{2}}$
$s = \sqrt{4 + 9 + 16}$
$s = \sqrt{29}$

the formula to compute t

$s = {v}_{0} t + \frac{1}{2} a {t}^{2}$
$\sqrt{29} = \left(0\right) \cdot t + \frac{1}{2} \left(\frac{7}{4}\right) {t}^{2}$
${t}^{2} = \frac{8}{7} \sqrt{29}$
$t = {\left(\frac{8}{7} \sqrt{29}\right)}^{\frac{1}{2}}$

$t = 2.480821248 \text{ }$seconds

God bless....I hope the explanation useful