An object is at rest at $(2, 8, 6)$ $(2 ,8 ,6 )$ and constantly accelerates at a rate of $\frac{5}{3} \frac{m}{s^{2}}$ $5/3 m/s^2$ as it moves to point B. If point B is at $(7, 5, 3)$ $(7 ,5 ,3 )$ , 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 · 2017-05-07T06:02:13

The time is $= 2.81 s$ $=2.81s$

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

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

$d = A B = \sqrt{{(7 - 2)}^{2} + {(5 - 8)}^{2} + {(3 - 6)}^{2}}$ $d=AB= sqrt((7-2)^2+(5-8)^2+(3-6)^2)$

$= \sqrt{5^{2} + 3^{2} + 3^{2}}$ $=sqrt(5^2+3^2+3^2)$

$= \sqrt{25 + 9 + 9}$ $=sqrt(25+9+9)$

$= \sqrt{43}$ $=sqrt43$

$= 6.56$ $=6.56$

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,

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

$t^{2} = \frac{2 d}{a} = \frac{2 \cdot 6.56}{\frac{5}{3}}$ $t^2=(2d)/a=(2*6.56)/(5/3)$

$= 7.87$ $=7.87$

$t = \sqrt{7.87} = 2.81 s$ $t=sqrt(7.87)=2.81s$

An object is at rest at (2 ,8 ,6 )(2,8,6)(2 ,8 ,6 ) and constantly accelerates at a rate of 5/3 m/s^253ms25/3 m/s^2 as it moves to point B. If point B is at (7 ,5 ,3 )(7,5,3)(7 ,5 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.