An object is at rest at (8 ,2 ,5 ) and constantly accelerates at a rate of 7/4 m/s^2 as it moves to point B. If point B is at (2 ,3 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Mar 18, 2017

The time taken is $= 2.71 s$

Explanation:

We are going to apply the following equation of motion :

$s = {u}_{0} t + \frac{1}{2} a {t}^{2}$

${u}_{0} = 0$

So,

$s = \frac{1}{2} a {t}^{2}$

$a = \frac{7}{4} m {s}^{-} 2$

The distance between 2 points $A = \left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $B = \left({x}_{2.} {y}_{2.} {z}_{2}\right)$ is

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

Here,

$A = \left(8 , 2 , 5\right)$ and $B = \left(2 , 3 , 7\right)$

So,

$s = \sqrt{{\left(2 - 8\right)}^{2} + {\left(3 - 2\right)}^{2} + {\left(7 - 5\right)}^{2}}$

$= \sqrt{36 + 1 + 4}$

$= \sqrt{41}$

From the equation of motion,

${t}^{2} = \frac{2 s}{a}$

${t}^{2} = \frac{2 \cdot \sqrt{41}}{\frac{7}{4}} = 7.32$

$t = \sqrt{7.32} = 2.71 s$