An object is at rest at $(4, 5, 1)$ $(4 ,5 ,1 )$ and constantly accelerates at a rate of $\frac{5}{3} \frac{m}{s^{2}}$ $5/3 m/s^2$ as it moves to point B. If point B is at $(7, 5, 3)$ $(7 ,5 ,3 )$ , 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 $(4, 5, 1)$ $(4 ,5 ,1 )$ and constantly accelerates at a rate of $\frac{5}{3} \frac{m}{s^{2}}$ $5/3 m/s^2$ as it moves to point B. If point B is at $(7, 5, 3)$ $(7 ,5 ,3 )$ , 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 · 2016-11-17T18:12:35

$t = 2.08$ $t=2.08$ s

Explanation:

$s^{2} = {(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2} + {(z_{2} - z_{1})}_{2}^{2}$ $s^2=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2$
$s^{2} = 13$ $s^2=13$
$s = \sqrt{13}$ $s=sqrt(13)$

An object is at rest $v_{0} = 0$ $v_0=0$
$s = v_{o} t + a \frac{t^{2}}{2} = a \frac{t^{2}}{2}$ $s=v_ot+at^2/2=at^2/2$
$t = \sqrt{2 \frac{s}{a}}$ $t=sqrt(2s/a)$
$t = \sqrt{2 \cdot \frac{\sqrt{13}}{5} \cdot 3}$ $t=sqrt(2*sqrt(13)/5*3$
$t = 2.08$ $t=2.08$ s

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An object is at rest at $(4, 5, 1)$ $(4 ,5 ,1 )$ and constantly accelerates at a rate of $\frac{5}{3} \frac{m}{s^{2}}$ $5/3 m/s^2$ as it moves to point B. If point B is at $(7, 5, 3)$ $(7 ,5 ,3 )$ , 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 (4,5,1)(4 ,5 ,1 ) and constantly accelerates at a rate of 53ms25/3 m/s^2 as it moves to point B. If point B is at (7,5,3)(7 ,5 ,3 ), 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 $(4, 5, 1)$ $(4 ,5 ,1 )$ and constantly accelerates at a rate of $\frac{5}{3} \frac{m}{s^{2}}$ $5/3 m/s^2$ as it moves to point B. If point B is at $(7, 5, 3)$ $(7 ,5 ,3 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.