An object is at rest at $(2, 1, 3)$ $(2 ,1 ,3 )$ and constantly accelerates at a rate of $1 \frac{m}{s^{2}}$ $1 m/s^2$ as it moves to point B. If point B is at $(3, 7, 5)$ $(3 ,7 ,5 )$ , 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 $(2, 1, 3)$ $(2 ,1 ,3 )$ and constantly accelerates at a rate of $1 \frac{m}{s^{2}}$ $1 m/s^2$ as it moves to point B. If point B is at $(3, 7, 5)$ $(3 ,7 ,5 )$ , 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 · 2018-04-04T06:11:58

The time is $= 3.58 s$ $=3.58s$

Explanation:

The distance $A B$ $AB$ is

$A B = \sqrt{{(3 - 2)}^{2} + {(7 - 1)}^{2} + {(5 - 3)}^{2}} = \sqrt{1 + 36 + 4} = \sqrt{41} m$ $AB=sqrt((3-2)^2+(7-1)^2+(5-3)^2)=sqrt(1+36+4)=sqrt(41)m$

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 $a = 1 m s^{- 2}$ $a=1ms^-2$

Therefore,

$\sqrt{41} = 0 + \frac{1}{2} \cdot 1 \cdot t^{2}$ $sqrt41=0+1/2*1*t^2$

$t^{2} = 2 \cdot \sqrt{41}$ $t^2=2*sqrt41$

$t = \sqrt{2 \cdot \sqrt{41}} = 3.58 s$ $t=sqrt(2*sqrt41)=3.58s$

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

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Explanation:

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1 Answer

Explanation:

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