An object is at rest at $(2, 4, 1)$ $(2 ,4 ,1 )$ and constantly accelerates at a rate of $\frac{2}{5} \frac{m}{s}$ $2/5 m/s$ as it moves to point B. If point B is at $(6, 6, 7)$ $(6 ,6 ,7 )$ , 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 · 2017-05-08T12:11:57

The time is $= 6.12 s$ $=6.12s$

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

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

$d = A B = \sqrt{{(6 - 2)}^{2} + {(6 - 4)}^{2} + {(7 - 1)}^{2}}$ $d=AB= sqrt((6-2)^2+(6-4)^2+(7-1)^2)$

$= \sqrt{4^{2} + 2^{2} + 6^{2}}$ $=sqrt(4^2+2^2+6^2)$

$= \sqrt{16 + 4 + 36}$ $=sqrt(16+4+36)$

$= \sqrt{56}$ $=sqrt56$

$= 7.48$ $=7.48$

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,

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

$t^{2} = \frac{2 d}{a} = \frac{2 \cdot 7.48}{\frac{2}{5}}$ $t^2=(2d)/a=(2*7.48)/(2/5)$

$= 37.42$ $=37.42$

$t = \sqrt{37.42} = 6.12 s$ $t=sqrt(37.42)=6.12s$

An object is at rest at (2 ,4 ,1 )(2,4,1)(2 ,4 ,1 ) and constantly accelerates at a rate of 2/5 m/s25ms2/5 m/s as it moves to point B. If point B is at (6 ,6 ,7 )(6,6,7)(6 ,6 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.