An object is at rest at $(1, 3, 9)$ $(1 ,3 ,9 )$ and constantly accelerates at a rate of $\frac{7}{4}$ $7/4$ $m s^{- 2}$ $ms^-2$ as it moves to point B. If point B is at $(2, 5, 6)$ $(2 ,5 ,6 )$ , 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 $(1, 3, 9)$ $(1 ,3 ,9 )$ and constantly accelerates at a rate of $\frac{7}{4}$ $7/4$ $m s^{- 2}$ $ms^-2$ as it moves to point B. If point B is at $(2, 5, 6)$ $(2 ,5 ,6 )$ , 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 · 2016-03-23T00:23:34

The distance between the points (calculated below) is $3.74$ $3.74$ $m$ $m$ . Then $d = u t + \frac{1}{2} a t^{2} = 0 (t) + \frac{1}{2} (\frac{7}{4}) t^{2} .$ $d=ut+1/2at^2=0(t)+1/2(7/4)t^2.$ Rearranging, $\frac{7}{8} t^{2} = 3.74 \to t = 2.1 s$ $7/8t^2=3.74 to t = 2.1 s$ .

Explanation:

To find the distance between the points:

$d = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2} + {(z_{2} - z_{1})}_{2}^{2}}$ $d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)$
$= \sqrt{{(2 - 1)}^{2} + {(5 - 3)}^{2} + {(6 - 9)}^{2}}$ $=sqrt((2-1)^2+(5-3)^2+(6-9)^2)$
$= \sqrt{1 + 4 + 9} = \sqrt{14} \approx 3.74$ $=sqrt(1+4+9)=sqrt14~~3.74$ $m$ $m$ .

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An object is at rest at $(1, 3, 9)$ $(1 ,3 ,9 )$ and constantly accelerates at a rate of $\frac{7}{4}$ $7/4$ $m s^{- 2}$ $ms^-2$ as it moves to point B. If point B is at $(2, 5, 6)$ $(2 ,5 ,6 )$ , 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 $(1, 3, 9)$ $(1 ,3 ,9 )$ and constantly accelerates at a rate of $\frac{7}{4}$ $7/4$ $m s^{- 2}$ $ms^-2$ as it moves to point B. If point B is at $(2, 5, 6)$ $(2 ,5 ,6 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.