An object is at rest at (8 ,6 ,2 ) and constantly accelerates at a rate of 1/4 m/s^2 as it moves to point B. If point B is at (2 ,8 ,2 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jul 1, 2016

It will take $7.11$ 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 $\left(8 , 6 , 2\right)$ and $\left(2 , 8 , 2\right)$ is

$\sqrt{{\left(2 - 8\right)}^{2} + {\left(8 - 6\right)}^{2} + {\left(2 - 2\right)}^{2}}$

= $\sqrt{{6}^{2} + {2}^{2} + {0}^{2}} = \sqrt{36 + 4 + 0} = \sqrt{40}$

(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 $\frac{1}{4}$ $\frac{m}{\sec} ^ 2$ will be given by

$t = \sqrt{\frac{2 \times \sqrt{40}}{\frac{1}{4}}} = \sqrt{4 \times 2 \times 6.3246} = \sqrt{17.888} = 7.11$