# An object is at rest at (4 ,5 ,1 ) and constantly accelerates at a rate of 4/3 m/s^2 as it moves to point B. If point B is at (7 ,2 ,6 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jun 18, 2016

#### Answer:

It takes $3.14$ s.

#### Explanation:

The equation of motion for a constant accelerating object is

$s = \frac{1}{2} a {t}^{2}$ where $s$ is the space traveled, $a$ is the acceleration and $t$ is the time.

We are interested in the time that is

$t = \sqrt{\frac{2 s}{a}}$.

The acceleration is given, then we need the space.
The distance between two point in space ${p}_{1} \left({x}_{1} , {y}_{1} , {z}_{1}\right) , {p}_{2} \left({x}_{2} , {y}_{2} , {z}_{2}\right)$ is given by

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

For us

$d = \sqrt{{\left(4 - 7\right)}^{2} + {\left(5 - 2\right)}^{2} + {\left(1 - 6\right)}^{2}}$

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

$= \sqrt{9 + 9 + 25} = \sqrt{43} \setminus \approx 6.56$.

This is the distance that the object has to travel, so we can write
$s = 6.56$ m. We now have all the ingredients

$t = \sqrt{\frac{2 s}{a}}$

$= \sqrt{\left(2 \cdot 6.56 {\text{ m")/(4/3"m"/"s}}^{2}\right)}$

=sqrt((13.12" m")/("m")*3/4 "s"^2)

$= \sqrt{9.84 {\text{ s}}^{2}}$

=sqrt(9.84" s"^2)\approx3.14" s".