An object is at rest at $(8, 3, 3)$ $(8 ,3 ,3 )$ and constantly accelerates at a rate of $\frac{1}{4} \frac{m}{s^{2}}$ $1/4 m/s^2$ as it moves to point B. If point B is at $(6, 3, 4)$ $(6 ,3 ,4 )$ , 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 $(8, 3, 3)$ $(8 ,3 ,3 )$ and constantly accelerates at a rate of $\frac{1}{4} \frac{m}{s^{2}}$ $1/4 m/s^2$ as it moves to point B. If point B is at $(6, 3, 4)$ $(6 ,3 ,4 )$ , 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 · 2017-09-14T09:26:52

The time is $= 8.46 s$ $=8.46s$

Explanation:

The distance $A B$ $AB$ is

$A B = \sqrt{{(6 - 8)}^{2} + {(3 - 3)}^{2} + {(4 - 3)}^{2}} = \sqrt{{(- 2)}^{2} + {(0)}^{2} + {(1)}^{2}}$ $AB=sqrt((6-8)^2+(3-3)^2+(4-3)^2)=sqrt((-2)^2+(0)^2+(1)^2)$

$= \sqrt{4 + 0 + 1} = \sqrt{5} m$ $=sqrt(4+0+1)=sqrt5m$

The acceleration is $a = \frac{1}{4} m s^{- 2}$ $a=1/4ms^-2$

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

We apply the equation of motion

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

So,

$\sqrt{5} = 0 + \frac{1}{2} \cdot {(\frac{1}{4})}^{2} \cdot t^{2}$ $sqrt5=0+1/2*(1/4)^2*t^2$

$t^{2} = 32 \sqrt{5}$ $t^2=32sqrt5$

$t = \sqrt{32 \sqrt{5}} = 8.46 s$ $t=sqrt(32sqrt5)=8.46s$

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An object is at rest at $(8, 3, 3)$ $(8 ,3 ,3 )$ and constantly accelerates at a rate of $\frac{1}{4} \frac{m}{s^{2}}$ $1/4 m/s^2$ as it moves to point B. If point B is at $(6, 3, 4)$ $(6 ,3 ,4 )$ , 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 $(8, 3, 3)$ $(8 ,3 ,3 )$ and constantly accelerates at a rate of $\frac{1}{4} \frac{m}{s^{2}}$ $1/4 m/s^2$ as it moves to point B. If point B is at $(6, 3, 4)$ $(6 ,3 ,4 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.