# An object is at rest at (3 ,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 (7 ,9 ,2 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Mar 17, 2017

The time taken is $= 2.94 s$

#### 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,

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

$a = \frac{4}{3} m {s}^{-} 2$

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,

$A = \left(3 , 5 , 1\right)$ and $B = \left(7 , 9 , 2\right)$

So,

$s = \sqrt{{\left(7 - 3\right)}^{2} + {\left(9 - 5\right)}^{2} + {\left(2 - 1\right)}^{2}}$

$= \sqrt{16 + 16 + 1}$

$= \sqrt{33}$

From the equation of motion,

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

${t}^{2} = \frac{2 \cdot \sqrt{33}}{\frac{4}{3}} = 8.62$

$t = \sqrt{8.62} = 2.94 s$