An object is at rest at (2 ,9 ,5 ) and constantly accelerates at a rate of 1/6 m/s as it moves to point B. If point B is at (6 ,2 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Apr 8, 2016

$\textcolor{b l u e}{\implies t = 9.984 \text{ seconds }}$ to 3 decimal places

Explanation:

$\textcolor{b l u e}{\text{Determine distance between points}}$

Let point start be ${P}_{s} \to \left({x}_{1} , {y}_{1} , {z}_{1}\right) \to \left(2 , 9 , 5\right)$
Let point end be ${P}_{e} \to \left({x}_{2} , {y}_{2} , {z}_{2}\right) \to \left(6 , 2 , 7\right)$

Let direct distance between ${P}_{s} \to {P}_{e}$ be $d$

Then by using Pythagoras we have:

$d = \sqrt{{\left[{x}_{2} - {x}_{1}\right]}^{2} + {\left[{y}_{2} - {y}_{1}\right]}^{2} + {\left[{z}_{2} - {z}_{1}\right]}^{2}}$

$d = \sqrt{{\left[6 - 2\right]}^{2} + {\left[2 - 9\right]}^{2} + {\left[7 - 5\right]}^{2}}$

$\textcolor{b l u e}{d = \sqrt{69} \text{ metres}}$
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Using the standard form equation for distance

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

Where
distance $\to s = \sqrt{69}$
initial velocity$\to u = 0$
acceleration $\to a = \frac{1}{6}$

$\implies \sqrt{69} = \left(\frac{1}{2}\right) \left(\frac{1}{6}\right) {t}^{2}$

$\implies {t}^{2} = 12 \sqrt{69}$

$\textcolor{b l u e}{\implies t = 9.984 \text{ seconds }}$ to 3 decimal places