# An object is at rest at (3 ,8 ,8 ) and constantly accelerates at a rate of 7/5 m/s^2 as it moves to point B. If point B is at (2 ,9 ,1 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jun 18, 2017

The time is $= 3.2 s$

#### 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

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

$d = A B = \sqrt{{\left(2 - 3\right)}^{2} + {\left(9 - 8\right)}^{2} + {\left(1 - 8\right)}^{2}}$

$= \sqrt{{1}^{2} + {1}^{2} + {7}^{2}}$

$= \sqrt{1 + 1 + 49}$

$= \sqrt{51}$

$= 7.14 m$

We apply the equation of motion

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

$u = 0$

so,

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

$a = \frac{7}{5} m {s}^{-} 2$

${t}^{2} = \frac{2 d}{a} = \frac{2 \cdot 7.14}{\frac{7}{5}}$

$= 10.2 {s}^{2}$

$t = \sqrt{10.2} = 3.2 s$