An object is at rest at $(5, 2, 1)$ $(5 ,2 ,1 )$ 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 $(6, 9, 2)$ $(6 ,9 ,2 )$ , 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 · 2018-07-03T06:22:13

The time is $= 5.98 s$ $=5.98s$

The distance $A B$ $AB$ is

$d_{A B} = \sqrt{{(6 - 5)}^{2} + {(9 - 2)}^{2} + {(2 - 1)}^{2}}$ $d_(AB)=sqrt((6-5)^2+(9-2)^2+(2-1)^2)$

$= \sqrt{1^{2} + 7^{2} + 1^{2}}$ $=sqrt(1^2+7^2+1^2)$

$= \sqrt{1 + 49 + 1}$ $=sqrt(1+49+1)$

$= \sqrt{51} m$ $=sqrt(51)m$

The acceleration is $a = \frac{2}{5} m s^{- 2}$ $a=2/5ms^-2$

The initial velocity $u = 0$ $u=0$

Apply the equation of motion

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

$\sqrt{51} = 0 + \frac{1}{2} \cdot \frac{2}{5} \cdot t^{2}$ $sqrt(51)=0+1/2*2/5*t^2$

$t^{2} = 5 \sqrt{51}$ $t^2=5sqrt(51)$

$t = \sqrt{5 \sqrt{51}} = 5.98 s$ $t=sqrt(5sqrt(51))=5.98s$

The time is $= 5.98 s$ $=5.98s$

An object is at rest at (5 ,2 ,1 )(5,2,1)(5 ,2 ,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 ,9 ,2 )(6,9,2)(6 ,9 ,2 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.