An object is at rest at $(8, 5, 4)$ $(8 ,5 ,4 )$ 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 $(2, 4, 5)$ $(2 ,4 ,5 )$ , 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 $(8, 5, 4)$ $(8 ,5 ,4 )$ 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 $(2, 4, 5)$ $(2 ,4 ,5 )$ , 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 · 2018-04-10T06:22:26

The time is $= 7.02 s$ $=7.02s$

Explanation:

The distance $A B$ $AB$ is

$A B = \sqrt{{(2 - 8)}^{2} + {(4 - 5)}^{2} + {(5 - 4)}^{2}} = \sqrt{36 + 1 + 1} = \sqrt{38} m$ $AB=sqrt((2-8)^2+(4-5)^2+(5-4)^2)=sqrt(36+1+1)=sqrt(38)m$

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

Therefore,

$\sqrt{38} = 0 + \frac{1}{2} \cdot \frac{1}{4} \cdot t^{2}$ $sqrt38=0+1/2*1/4*t^2$

$t^{2} = 8 \cdot \sqrt{38}$ $t^2=8*sqrt38$

$t = \sqrt{8 \cdot \sqrt{38}} = 7.02 s$ $t=sqrt(8*sqrt38)=7.02s$

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An object is at rest at $(8, 5, 4)$ $(8 ,5 ,4 )$ 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 $(2, 4, 5)$ $(2 ,4 ,5 )$ , 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 $(8, 5, 4)$ $(8 ,5 ,4 )$ 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 $(2, 4, 5)$ $(2 ,4 ,5 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.