# An object is at rest at (6 ,7 ,2 ) and constantly accelerates at a rate of 4/3 m/s^2 as it moves to point B. If point B is at (3 ,1 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Mar 27, 2018

$t = 3.24$

#### Explanation:

You can use the formula $s = u t + \frac{1}{2} \left(a {t}^{2}\right)$
$u$ is initial velocity
$s$ is distance travelled
$t$ is time
$a$ is acceleration

Now, it starts from rest so initial velocity is 0

$s = \frac{1}{2} \left(a {t}^{2}\right)$

To find s between $\left(6 , 7 , 2\right)$ and $\left(3 , 1 , 4\right)$
We use distance formula
$s = \sqrt{{\left(6 - 3\right)}^{2} + {\left(7 - 1\right)}^{2} + {\left(2 - 4\right)}^{2}}$
$s = \sqrt{9 + 36 + 4}$
$s = 7$

Acceleration is $\frac{4}{3}$ meters per second per second

$7 = \frac{1}{2} \left(\left(\frac{4}{3}\right) {t}^{2}\right)$
$14 \cdot \left(\frac{3}{4}\right) = {t}^{2}$
$t = \sqrt{10.5} = 3.24$