An object is at rest at (7 ,6 ,4 ) and constantly accelerates at a rate of 5/4 m/s^2 as it moves to point B. If point B is at (9 ,5 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jun 23, 2016

It will take $2.45$ 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 $\left(7 , 6 , 4\right)$ and $\left(9 , 5 , 7\right)$ is

$\sqrt{{\left(9 - 7\right)}^{2} + {\left(5 - 6\right)}^{2} + {\left(7 - 4\right)}^{2}}$

= $\sqrt{{2}^{2} + {1}^{2} + {3}^{2}} = \sqrt{4 + 1 + 9} = \sqrt{14}$

(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 $\frac{5}{4}$ $\frac{m}{\sec} ^ 2$ will be given by

t=sqrt((2sqrt14)/(5/4))=sqrt((2sqrt14)xx(4/5))=sqrt((8sqrt14)/5))

= $\sqrt{8 \times \frac{3.74166}{5}} = \sqrt{5.986656} = 2.45$