# An object is at rest at (1 ,3 ,5 ) and constantly accelerates at a rate of 4/3 m/s^2 as it moves to point B. If point B is at (1 ,9 ,8 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Mar 20, 2016

$t = 3 , 17 \text{ s}$

#### Explanation:

${P}_{A} = \left(\textcolor{red}{1} , \textcolor{g r e e n}{3} , \textcolor{b l u e}{5}\right) \text{ the point of A}$
${P}_{B} = \left(\textcolor{red}{1} , \textcolor{g r e e n}{9} , \textcolor{b l u e}{8}\right) \text{ the point of B}$
$\Delta x = \textcolor{red}{1 - 1 = 0}$
$\Delta y = \textcolor{g r e e n}{9 - 3 = 6}$
$\Delta z = \textcolor{b l u e}{8 - 5 = 3}$
$\Delta s : \text{distance between the " P_A" and } {P}_{B}$

$\Delta s = \sqrt{\Delta {x}^{2} + \Delta {y}^{2} + \Delta {z}^{2}}$
$\Delta s = \sqrt{\textcolor{red}{{0}^{2}} + \textcolor{g r e e n}{{6}^{2}} + \textcolor{b l u e}{{3}^{2}}}$
$\Delta s = \sqrt{\textcolor{g r e e n}{36} + \textcolor{b l u e}{9}}$
$\Delta s = \sqrt{45} \text{ "Delta s=6,71" m}$
$\text{use the equation below:}$
$\Delta s = \frac{1}{2} \cdot a \cdot {t}^{2} \text{ a=} \frac{4}{3} \frac{m}{s} ^ 2$
$6 , 71 = \frac{1}{2} \cdot \frac{4}{3} \cdot {t}^{2}$

${t}^{2} = \frac{6 \cdot 6 , 71}{4}$
${t}^{2} = 10 , 07$

$t = 3 , 17 \text{ s}$