# An object is at rest at (4 ,1 ,6 ) and constantly accelerates at a rate of 4/3 m/s^2 as it moves to point B. If point B is at (3 ,5 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Apr 8, 2016

#### Answer:

$t = 2 , 52 \text{ s}$

#### Explanation:

$\text{the point of A:(4,1,6)}$
$\text{the point of B:(3,5,7)}$

$\text{distance between the point A and B :}$
$s = \sqrt{{\left({A}_{x} - {B}_{x}\right)}^{2} + {\left({A}_{y} - {B}_{y}\right)}^{2} + {\left({A}_{z} - {B}_{z}\right)}^{2}}$

$s = \sqrt{{\left(4 - 3\right)}^{2} + {\left(1 - 5\right)}^{2} + {\left(6 - 7\right)}^{2}}$

$s = \sqrt{{1}^{2} + {\left(- 4\right)}^{2} + {\left(- 1\right)}^{2}}$

$s = \sqrt{1 + 16 + 1} = \sqrt{18}$

$s = 4 , 24 \text{ m}$

$\text{distance equation for an object starting from rest:}$

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

$\text{a:acceleration of object}$

$\text{t:elapsed time}$
$\text{given } a = \frac{4}{3} \frac{m}{s} ^ 2$

$4 , 24 = \frac{1}{2} \cdot \frac{4}{3} \cdot {t}^{2}$

${t}^{2} = \frac{6 \cdot 4 , 24}{4}$

${t}^{2} = 6 , 36$

$t = 2 , 52 \text{ s}$