# An object is at rest at (2 ,1 ,5 ) and constantly accelerates at a rate of 3 m/s as it moves to point B. If point B is at (6 ,7 ,5 ), 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 $2.193$ 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 $\left(2 , 1 , 5\right)$ and $\left(6 , 7 , 5\right)$ is

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

= $\sqrt{{4}^{2} + {6}^{2} + {0}^{2}} = \sqrt{16 + 36 + 0} = \sqrt{52} = 2 \sqrt{13}$

(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 $3$ $\frac{m}{\sec} ^ 2$ will be given by

$t = \sqrt{\frac{2 \times 2 \sqrt{13}}{3}} = \sqrt{\frac{4 \times 3.606}{3}} = \sqrt{4.808} = 2.193$