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

Jul 20, 2017

The time is $= 6.39 s$

#### Explanation:

The distance is

$s = \sqrt{{\left(6 - 2\right)}^{2} + {\left(4 - 1\right)}^{2} + {\left(5 - 6\right)}^{2}}$

$= \sqrt{16 + 9 + 1}$

$= \sqrt{26} m$

The initial velocity is $u = 0 m {s}^{-} 1$

The acceleration is $= \frac{1}{4} m {s}^{-} 2$

We apply the equation of motion

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

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

$t = \sqrt{\frac{2 \cdot \sqrt{26}}{\frac{1}{4}}}$

$t = 6.39 s$

Jul 20, 2017

$6.387$ seconds

#### Explanation:

The distance between $\left(2 , 1 , 6\right)$ and 6,4,5) is

$\sqrt{{\left(6 - 2\right)}^{2} + {\left(4 - 1\right)}^{2} + {\left(5 - 6\right)}^{2}} = \sqrt{16 + 9 + 1} = \sqrt{26}$, and it is in meters.

As distance $S = \frac{1}{2} a {t}^{2}$, where $a$ is accelaration and $t$ is time taken, then

$t = \sqrt{\frac{2 S}{a}} = \sqrt{\frac{2 \sqrt{26}}{\frac{1}{4}}}$

= $\sqrt{8 \sqrt{26}} = 2 \sqrt{2 \times 5.09902} = 2 \sqrt{10.19804} = 6.387$