# An object is at rest at (5 ,1 ,1 ) and constantly accelerates at a rate of 2/5 m/s as it moves to point B. If point B is at (6 ,9 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Nov 5, 2016

$t \approx 7 s$

#### Explanation:

I am going to assume that there is a small typographical error in the units of the acceleration and it is $\frac{2}{5} \frac{m}{s} ^ 2$

The distance, d, from Point $A = \left(5 , 1 , 1\right)$ to point $B = \left(6 , 9 , 7\right)$ is

$d = \sqrt{{\left(6 - 5\right)}^{2} + {\left(9 - 1\right)}^{2} + {\left(7 - 1\right)}^{2}}$

$d = \sqrt{{\left(1\right)}^{2} + {\left(8\right)}^{2} + {\left(6\right)}^{2}}$

$d = \sqrt{1 + 64 + 36}$

$d = \sqrt{101} m$

The equation for distance traveled under constant acceleration is:

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

Substitute $\sqrt{101} m$ for d and $\frac{2}{5} \frac{m}{s} ^ 2$ for a, and then solve for t:

$\sqrt{101} m = \left(\frac{1}{2}\right) \left(\frac{2}{5} \frac{m}{s} ^ 2\right) {t}^{2}$

$5 \sqrt{101} {s}^{2} = {t}^{2}$

$t \approx \pm 7 s$

But negative time does not make sense, therefore, make it only positive:

$t \approx 7 s$