An object is at rest at $(8, 3, 8)$ $(8 ,3 ,8 )$ 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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An object is at rest at $(8, 3, 8)$ $(8 ,3 ,8 )$ 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 · 2017-12-15T07:40:40

The time is $= 2.75 s$ $=2.75s$

Explanation:

The distance $A B$ $AB$ is

$A B = \sqrt{{(6 - 8)}^{2} + {(5 - 3)}^{2} + {(2 - 8)}^{2}} = \sqrt{4 + 4 + 36} = (\sqrt{44}) m$ $AB=sqrt((6-8)^2+(5-3)^2+(2-8)^2)=sqrt(4+4+36)=(sqrt44)m$

Apply the equation of motion

$s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$

The initial velocity is $u = 0 m s^{- 1}$ $u=0ms^-1$

The acceleration is $a = \frac{7}{4} m s^{- 2}$ $a=7/4ms^-2$

Therefore,

$\sqrt{44} = 0 + \frac{1}{2} \cdot \frac{7}{4} \cdot t^{2}$ $sqrt44=0+1/2*7/4*t^2$

$t^{2} = \frac{8}{7} \sqrt{44}$ $t^2=8/7sqrt44$

$t = \sqrt{\frac{8}{7} \sqrt{44}} = 2.75 s$ $t=sqrt(8/7sqrt44)=2.75s$

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An object is at rest at $(8, 3, 8)$ $(8 ,3 ,8 )$ 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.

1 Answer

Explanation:

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1 Answer

Explanation:

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An object is at rest at $(8, 3, 8)$ $(8 ,3 ,8 )$ 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.