# 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 ,9 ,2 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

May 10, 2016

2.77s

#### Explanation:

The coordinate of initial position of the object is$\left({x}_{1} = 4 , {y}_{1} = 5 , {z}_{1} = 1\right)$

The coordinate of final position of the object is$\left({x}_{2} = 7 , {y}_{2} = 9 , {z}_{2} = 2\right)$

Initial velocity of the object$u = 0$
Acceleration $a = \frac{4}{3} \frac{m}{s} ^ 2$

Distance traversed $s = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
=sqrt((7-4)^2+(9-5)^2+(2-1)^2))
$= \sqrt{9 + 16 + 1} = \sqrt{26} = 5.1 m$
If the time required be t sec then
$s = u t + \frac{1}{2} \cdot a \cdot {t}^{2} = 0 \cdot t + \frac{1}{2} \cdot \frac{4}{3} \cdot {t}^{2}$$\implies 5.1 = \frac{2}{3} \cdot {t}^{2}$
$t = \sqrt{\frac{15.3}{2}} = 2.77 s$