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

Apr 16, 2016

$t \cong 4 \text{ s}$

Explanation:

$A = \left(1 , 5 , 1\right) \text{ the point of A}$

$B = \left(9 , 4 , 8\right) \text{ the point of B}$

${A}_{x} = 1 \text{ "A_y=5" } {A}_{z} = 1$

${B}_{x} = 9 \text{ "B_y=4" } {B}_{z} = 8$

$\text{let's calculate distance between the points A and B}$

$\Delta s = \sqrt{{\left({B}_{x} - {A}_{x}\right)}^{2} + {\left({B}_{y} - {A}_{y}\right)}^{2} + {\left({B}_{z} - {A}_{z}\right)}^{2}}$

$\Delta s = \sqrt{{\left(9 - 1\right)}^{2} + {\left(4 - 5\right)}^{2} + {\left(8 - 1\right)}^{2}}$

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

$\Delta s = \sqrt{64 + 1 + 49}$

$\Delta s = 10 , 68 \text{ m}$

$a = \frac{4}{3} \frac{m}{s} ^ 2 \text{acceleration of object}$

$t = \text{elapsed time from the point of A to the point of B}$

$\Delta s = \frac{1}{2} \cdot a \cdot {t}^{2}$
$\text{distance equation for an object moving from rest}$

$10 , 68 = \frac{1}{2} \cdot \frac{4}{3} \cdot {t}^{2}$

${t}^{2} = \frac{3 \cdot 10 , 68}{2}$

${t}^{2} = 16 , 02$

$t \cong 4 \text{ s}$