An object is at rest at $(8, 6, 9)$ $(8 ,6 ,9 )$ and constantly accelerates at a rate of $1 \frac{m}{s}$ $1 m/s$ 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 · 2017-06-11T06:29:00

The time is $= 3.64 s$ $=3.64s$

The distance between the points $A = (x_{A}, y_{A}, z_{A})$ $A=(x_A,y_A,z_A)$ and the point $B = (x_{B}, y_{B}, z_{B})$ $B=(x_B,y_B,z_B)$ is

$A B = \sqrt{{(x_{B} - x_{A})}_{B}^{2} + {(y_{B} - y_{A})}_{B}^{2} + {(z_{B} - z_{A})}_{B}^{2}}$ $AB=sqrt((x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2)$

$d = A B = \sqrt{{(6 - 8)}^{2} + {(4 - 6)}^{2} + {(3 - 9)}^{2}}$ $d=AB= sqrt((6-8)^2+(4-6)^2+(3-9)^2)$

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

$= \sqrt{4 + 4 + 36}$ $=sqrt(4+4+36)$

$= \sqrt{44}$ $=sqrt44$

$= 6.63 m$ $=6.63m$

We apply the equation of motion

$d = u t + \frac{1}{2} a t^{2}$ $d=ut+1/2at^2$

$u = 0$ $u=0$

so,

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

$a = \frac{5}{4} m s^{- 2}$ $a=5/4ms^-2$

$t^{2} = \frac{2 d}{a} = \frac{2 \cdot 6.63}{1}$ $t^2=(2d)/a=(2*6.63)/(1)$

$= 13.27 s^{2}$ $=13.27s^2$

$t = \sqrt{13.27} = 3.64 s$ $t=sqrt(13.27)=3.64s$

An object is at rest at (8 ,6 ,9 )(8,6,9)(8 ,6 ,9 ) and constantly accelerates at a rate of 1 m/s1ms1 m/s 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.