An object is at rest at $(7, 6, 4)$ $(7 ,6 ,4 )$ and constantly accelerates at a rate of $\frac{5}{4} \frac{m}{s^{2}}$ $5/4 m/s^2$ as it moves to point B. If point B is at $(5, 5, 7)$ $(5 ,5 ,7 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

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An object is at rest at $(7, 6, 4)$ $(7 ,6 ,4 )$ and constantly accelerates at a rate of $\frac{5}{4} \frac{m}{s^{2}}$ $5/4 m/s^2$ as it moves to point B. If point B is at $(5, 5, 7)$ $(5 ,5 ,7 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

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Answer 1 · 2016-07-31T08:55:35

$2.447$ $2.447" "$ seconds to 3 decimal places

Explanation:

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

Let point 1 be $P_{1} \to (x_{1}, y_{1}, z_{1}) = (7, 6, 4)$ $P_1->(x_1,y_1,z_1)=(7,6,4)$
Let point 2 be $P_{2} \to (x_{2}, y_{2}, z_{2}) = (5, 5, 7)$ $P_2->(x_2,y_2,z_2)=(5,5,7)$

$Determine distance between points$ $color(blue)("Determine distance between points")$

$s = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2} + {(z_{2} - z_{1})}_{2}^{2}}$ $s = sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)$

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

$s = \sqrt{14}$ $color(blue)(s=sqrt(14))$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine time of travel$ $color(blue)("Determine time of travel")$

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

Mean velocity is
$v = \frac{0 + a t}{2} = \frac{a t}{2} = \frac{5}{4} \times \frac{1}{2} \times t = \frac{5}{8} t$ $v=(0+at)/2" " =" " (at)/2" " =" " 5/4xx1/2xxt" " =" " 5/8t$

$s = v t = (\frac{5}{8} t) \times t = \frac{5}{8} t^{2}$ $color(red)(s)" "color(green)(=" "vt" ")color(blue)( =" "(5/8 t)xxt)" "color(magenta)(=" "5/8t^2)$

$This is where the \frac{1}{2} a t^{2} in the standard equation$ $color(brown)("This is where the "1/2at^2 " in the standard equation")$ $of distance comes from$ $color(brown)("of distance comes from")$

$s = \frac{5}{8} t^{t} \Rightarrow t^{2} = \frac{8}{5} \times \sqrt{14}$ $s=5/8t^t" " =>" " t^2=8/5xxsqrt(14)$

$\Rightarrow t = \pm \sqrt{\frac{8}{5} \times \sqrt{14}} \approx \pm 2.447 to 3 decimal places$ $color(blue)(=>t=+-sqrt(8/5xxsqrt(14))~~+-2.447" to 3 decimal places")$

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

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Foot note about the units for acceleration$ $color(magenta)("Foot note about the units for acceleration")$

$You can manipulate the units of measurement in the same way$ $color(brown)("You can manipulate the units of measurement in the same way")$ $you do the counts (values)$ $color(brown)("you do the counts (values)")$

Acceleration is changing velocity

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

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

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

The passing of time is $t s$ $t s$

So the velocity after $t$ $t$ seconds is

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

Separating the counts from the units of measurement we have

$(a \times t) \times (\frac{m}{s^{2}} \times s)$ $(axxt) xx (m/s^2xxs)$

$\Rightarrow (a \times t) \times (\frac{m}{s} \times \frac{s}{s}) = (a t) \times (\frac{m}{s} \times 1) \to a t \frac{m}{s}$ $=>(axxt)xx(m/sxxs/s) = (at)xx(m/sxx1)" "->" "at m/s$

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An object is at rest at $(7, 6, 4)$ $(7 ,6 ,4 )$ and constantly accelerates at a rate of $\frac{5}{4} \frac{m}{s^{2}}$ $5/4 m/s^2$ as it moves to point B. If point B is at $(5, 5, 7)$ $(5 ,5 ,7 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer

Explanation:

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An object is at rest at (7,6,4)(7 ,6 ,4 ) and constantly accelerates at a rate of 54ms25/4 m/s^2 as it moves to point B. If point B is at (5,5,7)(5 ,5 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer

Explanation:

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An object is at rest at $(7, 6, 4)$ $(7 ,6 ,4 )$ and constantly accelerates at a rate of $\frac{5}{4} \frac{m}{s^{2}}$ $5/4 m/s^2$ as it moves to point B. If point B is at $(5, 5, 7)$ $(5 ,5 ,7 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.