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 (5 ,5 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jul 31, 2016

$2.447 \text{ }$ seconds to 3 decimal places

Explanation:

Let distance be $s$
Let acceleration be $a$
Let mean velocity be $v$

Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1} , {z}_{1}\right) = \left(7 , 6 , 4\right)$
Let point 2 be ${P}_{2} \to \left({x}_{2} , {y}_{2} , {z}_{2}\right) = \left(5 , 5 , 7\right)$

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

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

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

$\textcolor{b l u e}{s = \sqrt{14}}$
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$\textcolor{b l u e}{\text{Determine time of travel}}$

Distance is velocity multiplied by time but we have acceleration. So we need to change the acceleration to mean velocity

Mean velocity is
$v = \frac{0 + a t}{2} \text{ " =" " (at)/2" " =" " 5/4xx1/2xxt" " =" } \frac{5}{8} t$

color(red)(s)" "color(green)(=" "vt" ")color(blue)( =" "(5/8 t)xxt)" "color(magenta)(=" "5/8t^2)

$\textcolor{b r o w n}{\text{This is where the "1/2at^2 " in the standard equation}}$$\textcolor{b r o w n}{\text{of distance comes from}}$

$s = \frac{5}{8} {t}^{t} \text{ " =>" } {t}^{2} = \frac{8}{5} \times \sqrt{14}$

$\textcolor{b l u e}{\implies t = \pm \sqrt{\frac{8}{5} \times \sqrt{14}} \approx \pm 2.447 \text{ to 3 decimal places}}$

The negative time is not logical so we can discount it.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{m a \ge n t a}{\text{Foot note about the units for acceleration}}$

$\textcolor{b r o w n}{\text{You can manipulate the units of measurement in the same way}}$$\textcolor{b r o w n}{\text{you do the counts (values)}}$

Acceleration is changing velocity

Let the value of the acceleration is $a$
Let the unit measure of distance be $m$
Let the unit measure of time be $s$
Let the count of time be $t$

Then acceleration is written as $a \frac{m}{s} ^ 2$

We must write the seconds as ${s}^{2}$ to make what follows work.

The passing of time is $t s$

So the velocity after $t$ seconds is

$a \frac{m}{s} ^ 2 \times t s$

Separating the counts from the units of measurement we have

$\left(a \times t\right) \times \left(\frac{m}{s} ^ 2 \times s\right)$

$\implies \left(a \times t\right) \times \left(\frac{m}{s} \times \frac{s}{s}\right) = \left(a t\right) \times \left(\frac{m}{s} \times 1\right) \text{ "->" } a t \frac{m}{s}$