An object is at rest at $(6, 2, 8)$ $(6 ,2 ,8 )$ and constantly accelerates at a rate of $1 \frac{m}{s^{2}}$ $1 m/s^2$ as it moves to point B. If point B is at $(3, 7, 5)$ $(3 ,7 ,5 )$ , 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 · 2017-06-17T06:11:39

The time is $= 3.62 s$ $=3.62s$

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

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

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

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

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

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

$= 6.56 m$ $=6.56m$

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,

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

$a = 1 m s^{- 2}$ $a=1ms^-2$

$t^{2} = \frac{2 d}{a} = \frac{2 \cdot 6.56}{1}$ $t^2=(2d)/a=(2*6.56)/(1)$

$= 13.11 s^{2}$ $=13.11s^2$

$t = \sqrt{13.11} = 3.62 s$ $t=sqrt(13.11)=3.62s$

An object is at rest at (6,2,8)(6 ,2 ,8 ) and constantly accelerates at a rate of 1ms21 m/s^2 as it moves to point B. If point B is at (3,7,5)(3 ,7 ,5 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.