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

Mar 27, 2017

The time is $= 3.89 s$

#### Explanation:

The distance between the 2 points $A = \left(2 , 1 , 3\right)$ and $B = \left(9 , 3 , 5\right)$ is

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

$= \sqrt{49 + 4 + 4}$

$= \sqrt{57}$

The acceleration is

$a \left(t\right) = 1$

The velocity is

$v \left(t\right) = {\int}_{0}^{t} 1 \mathrm{dt} = t$

The distance is

$s = {\int}_{0}^{t} t \mathrm{dt} = \frac{1}{2} {t}^{2}$

This is equal to

sqrt57=1/2t^2

${t}^{2} = 2 \sqrt{57} = 15.1$

So,

$t = \sqrt{15.1} = 3.89 s$