An object is at rest at (4 ,8 ,3 ) 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 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Mar 27, 2016

$t = 3 , 49 \text{ s}$

Explanation:

$\text{The Point of A=} \left(4 , 8 , 3\right)$
$\text{The Point of B=} \left(3 , 1 , 7\right)$
$\text{distance between two point:}$
$s = \sqrt{{\left({B}_{x} - {A}_{x}\right)}^{2} + {\left({B}_{y} - {A}_{y}\right)}^{2} + {\left({B}_{z} - {A}_{z}\right)}^{2}}$

$s = \sqrt{{\left(3 - 4\right)}^{2} + {\left(1 - 8\right)}^{2} + {\left(7 - 3\right)}^{2}}$

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

$s = \sqrt{1 + 49 + 16} \text{ } s = \sqrt{66}$

$s = \frac{1}{2} \cdot a \cdot {t}^{2}$
$\text{equation for object moving at constant acceleration from rest}$
$\sqrt{66} = \frac{1}{2} \cdot \frac{4}{3} \cdot {t}^{2}$
$6 \cdot \sqrt{66} = 4 \cdot {t}^{2}$
$6 \cdot 8 , 12 = 4 \cdot {t}^{2}$
$48 , 72 = 4 \cdot {t}^{2}$
t²=(48,72)/4
${t}^{2} = 12 , 18$

$t = 3 , 49 \text{ s}$