An object is at rest at $(9, 7, 5)$ $(9 ,7 ,5 )$ and constantly accelerates at a rate of $\frac{5}{4} \frac{m}{s^{2}}$ $5/4 m/s^2$ as it moves to point B. If point B is at $(8, 2, 6)$ $(8 ,2 ,6 )$ , 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 · 2017-06-04T06:59:21

The time is $= 2.88 s$ $=2.88s$

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

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

$d = A B = \sqrt{{(8 - 9)}^{2} + {(2 - 7)}^{2} + {(6 - 5)}^{2}}$ $d=AB= sqrt((8-9)^2+(2-7)^2+(6-5)^2)$

$= \sqrt{1^{2} + 5^{2} + 1^{2}}$ $=sqrt(1^2+5^2+1^2)$

$= \sqrt{1 + 25 + 1}$ $=sqrt(1+25+1)$

$= \sqrt{27}$ $=sqrt27$

$= 5.196 m$ $=5.196m$

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,

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

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

$t^{2} = \frac{2 d}{a} = \frac{2 \cdot 5.196}{\frac{5}{4}}$ $t^2=(2d)/a=(2*5.196)/(5/4)$

$= 8.31 s^{2}$ $=8.31s^2$

$t = \sqrt{8.31} = 2.88 s$ $t=sqrt(8.31)=2.88s$

An object is at rest at (9 ,7 ,5 )(9,7,5)(9 ,7 ,5 ) and constantly accelerates at a rate of 5/4 m/s^254ms25/4 m/s^2 as it moves to point B. If point B is at (8 ,2 ,6 )(8,2,6)(8 ,2 ,6 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.