# An object is at rest at (7 ,4 ,2 ) and constantly accelerates at a rate of 4/3 m/s as it moves to point B. If point B is at (3 ,1 ,9 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Apr 27, 2016

${t}_{F} \approx 3.592 \text{ }$ to 3 decimal places

#### Explanation:

$\textcolor{red}{\text{Note that I have shown how to deal with units}}$

$\textcolor{g r e e n}{\text{Tip:- leave the converting to decimal until the end}}$

$\textcolor{b l u e}{\text{Determine the distance of travel}}$

Let distance be $d$
Let initial time be ${t}_{0}$ seconds
Let time at any instant $i$ be ${t}_{i}$ seconds
Let final time be ${t}_{F}$ seconds

Given: acceleration $\to \frac{4}{3} \textcolor{w h i t e}{.} \textcolor{red}{\frac{m}{s} ^ 2}$

Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1} , {z}_{1}\right) \to \left(7 , 4 , 2\right)$
Let point 2 be ${P}_{2} \to \left({x}_{2} , {y}_{2} , {z}_{2}\right) \to \left(3 , 1 , 9\right)$

Using Pythagoras

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

$d = \sqrt{{\left(3 - 7\right)}^{2} + {\left(1 - 4\right)}^{2} + {\left(9 - 2\right)}^{2}}$

$d = \sqrt{16 + 9 + 49}$

$d = \sqrt{74} \text{ }$ metres$\text{ }$(74 is a product of 2 primes)
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$\textcolor{b l u e}{\text{Determine time of travel}}$

Velocity at ${t}_{0} = 0 \textcolor{w h i t e}{.} \frac{m}{s}$

Velocity at ${t}_{i} = \frac{4}{3} \times {t}_{i} \textcolor{g r e e n}{\text{ units} \to \frac{m}{s} ^ s \times s \textcolor{b l u e}{\to m \times \frac{s}{s} ^ 2 = \frac{m}{s}}}$

Mean velocity $\frac{1}{2} \times \frac{4}{3} \times {t}_{F} \textcolor{w h i t e}{.} \textcolor{b l u e}{\frac{m}{s}}$

Thus distance is mean velocity times time giving

$\text{ } d = \frac{1}{2} \times \frac{4}{3} \times {t}_{F} \times {t}_{F}$

" "d= 4/6(t_F)^2 color(green)(" units"-> m/sxxs ->color(blue)(mxxs/s=m

Multiply both sides by $\frac{6}{4}$

$\text{ } d \frac{6}{4} = \frac{4}{6} \times \frac{6}{4} {\left({t}_{F}\right)}^{2}$

But $\frac{4}{6} \times \frac{6}{4} = 1$

$\text{ } \implies d \frac{6}{4} = {\left({t}_{F}\right)}^{2}$

Thus

$\text{ } {t}_{F} = \sqrt{\frac{6}{4}} \times \sqrt{\sqrt{74}}$
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Consider $\sqrt{\frac{6}{4}} \to \sqrt{\frac{6}{{2}^{2}}} = \frac{1}{2} \sqrt{6}$

Consider $\sqrt{\sqrt{74}} \to \sqrt{74}$
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$\textcolor{b r o w n}{\text{Putting it all together}}$

$\text{ } {t}_{F} = \frac{1}{2} \sqrt{6} \times \sqrt{74}$

By calculator: ${t}_{F} \approx 3.592 \text{ }$ to 3 decimal places