# An object is at rest at (8 ,3 ,8 ) and constantly accelerates at a rate of 7/4 m/s^2 as it moves to point B. If point B is at (6 ,3 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Mar 18, 2017

The time is $= 2.26 s$

#### Explanation:

We are going to apply the following equation of motion :

$s = {u}_{0} t + \frac{1}{2} a {t}^{2}$

${u}_{0} = 0$

So,

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

$a = \frac{7}{4} m {s}^{-} 2$

The distance between 2 points $A = \left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $B = \left({x}_{2.} {y}_{2.} {z}_{2}\right)$ is

$= \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$

Here,

$A = \left(8 , 3 , 8\right)$ and $B = \left(6 , 3 , 4\right)$

So,

$s = \sqrt{{\left(6 - 8\right)}^{2} + {\left(3 - 3\right)}^{2} + {\left(4 - 8\right)}^{2}}$

$= \sqrt{4 + 0 + 16}$

$= \sqrt{20}$

From the equation of motion,

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

${t}^{2} = \frac{2 \cdot \sqrt{20}}{\frac{7}{4}} = 5.11$

$t = \sqrt{5.11} = 2.26 s$