# An object is at rest at (7 ,8 ,4 ) and constantly accelerates at a rate of 5/4 m/s^2 as it moves to point B. If point B is at (1 ,5 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

May 28, 2016

It takes 3.29 seconds.

#### Explanation:

First of all we need to know how much distance the objects has to travel. The distance between two points is given by the Pitagora's theorem in 3 dimensions

$d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2} + {\left({z}_{1} - {z}_{2}\right)}^{2}}$

in our case

$d = \sqrt{{\left(7 - 1\right)}^{2} + {\left(8 - 5\right)}^{2} + {\left(4 - 3\right)}^{2}} = \sqrt{36 + 9 + 1} = \sqrt{46} \setminus \approx 6.78 m$

Now that we know that our body will travel for 6.78 m we can use the equation of motion for an accelerating body that is

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

where $x$ is the traveled distance, for us 6.78, $a$ is the acceleration, for us $\frac{5}{4} \frac{m}{s} ^ 2$ and $t$ is the time that we want to find. I plug the values in the equation

$6.78 = \frac{1}{2} \cdot \frac{5}{4} \cdot {t}^{2}$
$6.78 = \frac{5}{8} {t}^{2}$

I am interested in $t$ so I multiply both sides for $\frac{8}{5}$ and do the square root

$\sqrt{6.78 \cdot \frac{8}{5}} = t$
$t = \sqrt{10.848} \setminus \approx 3.29$ s.