# 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 (3 ,4 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Oct 13, 2016

It will take the object $5$ seconds to reach point B.

#### Explanation:

You can use the equation
$r = v \setminus \Delta t + \frac{1}{2} a \setminus \Delta {t}^{2}$
where $r$ is the separation between the two points, $v$ is the initial velocity (here $0$, as at rest), $a$ is acceleration and $\setminus \Delta t$ is the elapsed time (which is what you want to find).

The distance between the two points is
$\left(3 , 4 , 7\right) - \left(2 , 1 , 6\right) = \left(3 - 2 , 4 - 1 , 7 - 6\right) = \left(1 , 3 , 1\right)$

r = || (1,3,1) || = $\setminus \sqrt{{1}^{2} + {3}^{2} + {1}^{2}} = \setminus \sqrt{11} = 3.3166 \setminus \textrm{m}$

Substitute $r = 3.3166$, $a = \frac{1}{4}$ and $v = 0$ into the equation given above
$3.3166 = 0 + \frac{1}{2} \frac{1}{4} \setminus \Delta {t}^{2}$ Rearrange for $\setminus \Delta t$
$\setminus \Delta t = \setminus \sqrt{\left(8\right) \left(3.3166\right)}$
$\setminus \Delta t = 5.15 \setminus \textrm{s}$

Round to however many decimal places are requested, or to significant figures, of which here there is one, so $5 s$.