An object is at rest at $(3, 7, 2)$ $(3 ,7 ,2 )$ and constantly accelerates at a rate of $\frac{7}{5} \frac{m}{s^{2}}$ $7/5 m/s^2$ as it moves to point B. If point B is at $(5, 0, 9)$ $(5 ,0 ,9 )$ , 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 $(3, 7, 2)$ $(3 ,7 ,2 )$ and constantly accelerates at a rate of $\frac{7}{5} \frac{m}{s^{2}}$ $7/5 m/s^2$ as it moves to point B. If point B is at $(5, 0, 9)$ $(5 ,0 ,9 )$ , 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 · 2018-01-04T14:32:06

The time is $= 3.8 s$ $=3.8s$

Explanation:

The distance $A B$ $AB$ is

$A B = \sqrt{{(5 - 3)}^{2} + {(0 - 7)}^{2} + {(9 - 2)}^{2}} = \sqrt{4 + 49 + 49} = (\sqrt{102}) m$ $AB=sqrt((5-3)^2+(0-7)^2+(9-2)^2)=sqrt(4+49+49)=(sqrt102)m$

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 $a = \frac{7}{5} m s^{- 2}$ $a=7/5ms^-2$

Therefore,

$\sqrt{102} = 0 + \frac{1}{2} \cdot \frac{7}{5} \cdot t^{2}$ $sqrt(102)=0+1/2*7/5*t^2$

$t^{2} = \frac{10}{7} \sqrt{102}$ $t^2=10/7sqrt102$

$t = \sqrt{\frac{10}{7} \sqrt{102}} = 3.8 s$ $t=sqrt(10/7sqrt102)=3.8s$

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An object is at rest at $(3, 7, 2)$ $(3 ,7 ,2 )$ and constantly accelerates at a rate of $\frac{7}{5} \frac{m}{s^{2}}$ $7/5 m/s^2$ as it moves to point B. If point B is at $(5, 0, 9)$ $(5 ,0 ,9 )$ , 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 $(3, 7, 2)$ $(3 ,7 ,2 )$ and constantly accelerates at a rate of $\frac{7}{5} \frac{m}{s^{2}}$ $7/5 m/s^2$ as it moves to point B. If point B is at $(5, 0, 9)$ $(5 ,0 ,9 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.