# An object is at rest at (6 ,5 ,9 ) and constantly accelerates at a rate of 4/3 m/s^2 as it moves to point B. If point B is at (3 ,6 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jun 24, 2016

It will take $2.979$ seconds.

#### Explanation:

The distance between two points $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ is given by

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

Hence distance between $\left(6 , 5 , 9\right)$ and $\left(3 , 6 , 4\right)$ is

$\sqrt{{\left(3 - 6\right)}^{2} + {\left(6 - 5\right)}^{2} + {\left(4 - 9\right)}^{2}}$

= $\sqrt{{3}^{2} + {1}^{2} + {5}^{2}} = \sqrt{9 + 1 + 25} = \sqrt{35} = 5.916$

(As distance covered is given by $S = u t + \frac{1}{2} a {t}^{2}$, where $u$ is initial velocity, $a$ is accelaration and $t$ is time taken. If body is at rest $S = \frac{1}{2} a {t}^{2}$ and hence $t = \sqrt{\frac{2 S}{a}}$

As the coordinates are in meters, the time taken at an acceleration of $\frac{4}{3}$ $\frac{m}{\sec} ^ 2$ will be given by

$t = \sqrt{\frac{2 \times 5.916}{\frac{4}{3}}} = \sqrt{\frac{6 \times 5.916}{4}} = \sqrt{8.874} = 2.979$