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

Mar 19, 2016

$t = 3 , 44 \text{ s}$

#### Explanation:

$\text{distance between two points:}$
$\Delta x = 3 - 4 = - 1$
$\Delta y = 1 - 7 = - 6$
$\Delta z = 6 - 1 = 5$
$s = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} + {\left(\Delta z\right)}^{2}}$
$s = \sqrt{{\left(- 1\right)}^{2} + {\left(- 6\right)}^{2} + {5}^{2}}$
$s = \sqrt{1 + 36 + 25} \text{ "s=sqrt62" m}$
$s = 7 , 87 \text{ m}$

$s = \frac{1}{2} \cdot a {t}^{2}$
$7 , 87 = \frac{1}{\cancel{2}} \cdot \frac{\cancel{4}}{3} \cdot {t}^{2}$
${t}^{2} = \frac{7 , 87 \cdot 3}{2}$
${t}^{2} = 11 , 81$
$t = \sqrt{11 , 81}$
t=±3,44
$t = \cancel{- 3 , 44} \text{ there is no negative time}$
$t = 3 , 44 \text{ s}$

Mar 19, 2016

$t = \sqrt{3 \sqrt{\frac{31}{2}}} = 3.4367$

#### Explanation:

Given Location of an object,
Rest position $A \left(4 , 7 , 1\right)$
New Location $B \left(3 , 1 , 6\right)$
Acceleration, $a = \frac{4}{3} \frac{m}{s} ^ 2$
Find the time it take to move from $A \to B$
Using the distance formula we find the distance from $A \to B$
$\overline{A B} = r = \sqrt{{\left(4 - 3\right)}^{2} + {\left(7 - 1\right)}^{2} + {\left(1 - 6\right)}^{2}} = \sqrt{62}$
Now we need to use the kinematic equation of motion that uses
$t , a , r$ as variable.
F(r,t;v_o,a) => r = v_ot +1/2at^2 ; v_o = 0, r=sqrt(62); a=4/3m/s^2
$\sqrt{62} = \frac{1}{2} \cdot \frac{4}{3} {t}^{2}$
$t = \sqrt{3 \sqrt{\frac{31}{2}}} = 3.4367$