An object is at rest at $(1, 6, 9)$ $(1 ,6 ,9 )$ and constantly accelerates at a rate of $1$ $1$ $m s^{- 2}$ $ms^-2$ as it moves to point B. If point B is at $(6, 4, 3)$ $(6 ,4 ,3 )$ , 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 · 2017-06-07T12:33:57

The distance is $8.06$ $8.06$ $m$ $m$ , and the object will take $4.01$ $4.01$ $s$ $s$ to travel this distance.

First we need to find the distance between the two points:

$s = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2} + {(z_{2} - z_{1})}_{2}^{2}}$ $s=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)$

$s = \sqrt{{(6 - 1)}^{2} + {(4 - 6)}^{2} + {(3 - 9)}^{2}}$ $s=sqrt((6-1)^2+(4-6)^2+(3-9)^2)$

$s = \sqrt{{(5)}^{2} + {(- 2)}^{2} + {(- 6)}^{2}}$ $s=sqrt((5)^2+(-2)^2+(-6)^2)$

$s = \sqrt{25 + 4 + 36} = \sqrt{65} = 8.06$ $s=sqrt(25+4+36) = sqrt(65) = 8.06$ $m$ $m$ (assuming the units are in metres)

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

The initial velocity, $u$ $u$ , is equal to $0$ $0$ , so the first term disappears.

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

Rearranging:

$t = \sqrt{\frac{2 s}{a}} = \sqrt{\frac{2 (8.06)}{1}} = 4.01$ $t=sqrt((2s)/a)=sqrt((2(8.06))/1)=4.01$ $s$ $s$

An object is at rest at (1 ,6 ,9 )(1,6,9)(1 ,6 ,9 ) and constantly accelerates at a rate of 111 ms^-2ms−2 ms^-2 as it moves to point B. If point B is at (6 ,4 ,3 )(6,4,3)(6 ,4 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.