An object is at rest at $(8, 1, 5)$ $(8 ,1 ,5 )$ and constantly accelerates at a rate of $\frac{7}{4} \frac{m}{s^{2}}$ $7/4 m/s^2$ as it moves to point B. If point B is at $(6, 5, 2)$ $(6 ,5 ,2 )$ , 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 · 2016-04-05T15:30:16

$t = 2, 48 s$ $t=2,48" s"$

$first; let's find distance between the point of(8,1,5) and (6,5,2)$ $"first; let's find distance between the point of(8,1,5) and (6,5,2)"$

$s = \sqrt{{(6 - 8)}^{2} + {(5 - 1)}^{2} + {(2 - 5)}^{2}}$ $s=sqrt((6-8)^2+(5-1)^2+(2-5)^2)$

$s = \sqrt{{(- 2)}^{2} + 4^{2} + {(- 3)}^{2}} = \sqrt{4 + 16 + 9} = \sqrt{29} m$ $s=sqrt((-2)^2+4^2+(-3)^2)=sqrt(4+16+9)=sqrt(29)" m"$

$now ; let's use the equation of s = \frac{1}{2} \cdot a t^{2}$ $"now ; let's use the equation of "s=1/2*a t^2$

$\sqrt{29} = \frac{1}{2} \cdot \frac{7}{4} \cdot t^{2}$ $sqrt29=1/2*7/4* t^2$

$t^{2} = \frac{8 \cdot \sqrt{29}}{7}$ $t^2=(8*sqrt29)/7$

$t^{2} = \frac{8 \cdot 5, 39}{7}$ $t^2=(8*5,39)/7$

$t = 2, 48 s$ $t=2,48" s"$

An object is at rest at (8 ,1 ,5 )(8,1,5)(8 ,1 ,5 ) and constantly accelerates at a rate of 7/4 m/s^274ms27/4 m/s^2 as it moves to point B. If point B is at (6 ,5 ,2 )(6,5,2)(6 ,5 ,2 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.