# An object is at rest at (1 ,3 ,5 ) and constantly accelerates at a rate of 4/3 m/s^2 as it moves to point B. If point B is at (3 ,6 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jul 2, 2017

The time is $= 2.37 s$

#### Explanation:

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

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

$d = A B = \sqrt{{\left(3 - 1\right)}^{2} + {\left(6 - 3\right)}^{2} + {\left(4 - 5\right)}^{2}}$

$= \sqrt{{2}^{2} + {3}^{2} + {1}^{2}}$

$= \sqrt{4 + 9 + 1}$

$= \sqrt{14}$

$= 3.74 m$

We apply the equation of motion

$d = u t + \frac{1}{2} a {t}^{2}$

$u = 0$

so,

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

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

${t}^{2} = \frac{2 d}{a} = \frac{2 \cdot 3.74}{\frac{4}{3}}$

$= 5.61 {s}^{2}$

$t = \sqrt{5.61} = 2.37 s$